QUANTITY OF HEAT SIMPLIFIED PHYSICS NOTES

Chapter Nine

QUANTITY OF HEAT

• Heat is a form of energy that flows from one body to another due to a temperature difference between them.
• The absorption of heat by a body results in rise of its temperature while loss of the same results in fall of temperature.

HEAT CAPACITY

It takes shorter time to prepare tea for fewer people as compared to preparing tea for many.

This is because less amount of energy is needed to prepare tea for fewer people. On a hot day, the land surface is much warmer compared to the sea. This is because of the nature of the surfaces. Different materials have different rates of heat absorption.

EXPERIMENT: To investigate the relationship between the mass of a body and the quantity of heat required to cause a unit temperature rise in it

Apparatus Water, beaker, Bunsen burner, thermometer, stopwatch, wire gauze.

Procedure

• Heat about 150 cm3 of water at room temperature, as shown in figure 9.1
• Record the time taken for the temperature to rise to about 60 °C.
•  Pour out the water and cool the beaker to room temperature.
• Repeat the experiment with about 200 cm3 of water in the beaker.
• Record the time taken for the temperature to rise to about 60 °C.

Observation

It takes a longer time for the larger volume of water to attain the same temperature rise than for the smaller volume of water.

Explanation

• The different volumes are heated from the same initial temperature to final temperature.
• The larger volume takes more time to attain the same temperature change, hence absorbs more heat energy.

Conclusion

• Since the different volumes of water have different masses, the quantity of heat energy required to cause a given temperature change depends on its mass.

• Heat capacity is defined as the quantity of heat energy required to raise the temperature of a given mass of a material by one degree Celsius or one Kelvin. It is denoted by C.

The SI unit of heat capacity is JK-1

Example 1

Calculate the quantity of heat required to raise the temperature of a metal block with a heat capacity of 460 IK-1 from 15 °C to 45 °C.

Solution

Heat capacity, C = 460 J K-1

Temperature change 9 = (45°C — 15°C) = 30 °C

 Q = CΔT

= 460 x 30

= 13 800 J

 

SPECIFIC HEAT CAPACITY

• Specific heat capacity is defined as the quantity of heat required to raise the temperature of a unit mass of a substance by one Kelvin (K). It is denoted by c. Specific heat capacity is heat capacity per unit mass. ‘

c =             quantity of heat                                 (H)

Mass (m) x change in temperature (ΔT)

H = mc ΔT

• The SI unit for specific heat capacity is Jkg-1K-1.

Note:   If two different substances of the same mass are subjected to the same amount of heat, they acquire different temperature changes.

For example, the specific heat capacity of copper is 390 Jkg-1K-1. This means that 1 kg of copper would take in or give out 390 J when its temperature changes by 1 K.

Example 2

1. A block of metal of mass 1.5 kg which is suitably insulated is heated from 30 °C to 50 °C in

8 minutes and 20 seconds by an electric heater coil rated 54 watts. Find:

(a) the quantity of heat supplied by the heater.

(b) the heat capacity of the block.

(c) its specific heat capacity.

Solution

• Quantity of heat supplied = power x  time

Q = 54  x  500

= 27000J

(b) Heat capacity, C = Q/ΔT

But Q = 27000 J and ΔT= 50 – 30 =20K

C  = 27000

     20

= 1350 JK-1

(c)           Specific heat capacity, c = C/m
But c = 1 350 and m = 1.5

C = 1350

1.5

 

=900Jkg-1K-1

2. Find the temperature of water if a heater rated 42 W heat 50 g water from 20C in five minute. (Specific heat capacity of water is 4200Jkg-1K-1)

 

Assuming no heat losses;

Heat supplied by the heater = heat gained by the water.

42 x 5 x 60 = mcΔT

42 x 300 = 50 X 10-3 X 4 200 x ΔT

ΔT =       42 x 300

50 X 10-3 X 4 200

ΔT= 60

But ΔT= T – 20, where T is the final temperature.

60= T- 20
T = 60 + 20

T = 80°C

3. A piece of copper of mass 60 g and specific heat capacity 390 Jkg-1K-1cools from 90°C to 40°C Find the quantity of heat given out.

Solution

Q = mcΔT= 60 x 10^–3 x 390 x 50
= 1 170 J

 

DETERMINATION OF SPECIFIC HEAT CAPACITY

Method of Mixtures

Solids

EXPERIMENT: To determine the specific heat capacity of a solid by the method of mixtures

Apparatus

Metal block, thread, beaker, water, tripod stand, heat source, well-lagged calorimeter, stirrer, thermometer, cardboard.

 

 

Procedure

• Weigh the solid metal block. = ms
• Set up the apparatus as shown in figure above. Allow the water to boil.
• Weigh the calorimeter together with the stirrer and pour some water into it. = mc
• Weigh the calorimeter with its contents and place it in the insulating jacket. = m1
• Mass of water = m1- mc = mw
• Measure the temperature of the cold water. = T1
• Measure temperature of boiling water in beaker = T2
• When the water in the beaker has boiled for sometime, quickly transfer the metal block, from the beaker into the cold water in the calorimeter.
• Cover the calorimeter with a piece of cardboard, as in figure.
• Stir the mixture and record the final temperature. = T3
• Change of temperature of water in the calorimeter on addition of hot metal block = T3- T1
• Temperature change of hot metal block when dropped into the cold water = T2- T3,
  Assuming no heat losses to the surroundings during the transfer of the metal block from the beaker to the calorimeter and thereafter, the specific heat capacity of the solid can be calculated as follows;

Heat lost by           heat gained by                          gained by

metal block     =    calorimeter with stirrer   +       water in calorimeter

mscs (T2 – T3) =  mccc (T3 – T1) + mwcw (T3 – T1)

 

cs = m_cc_c(T₃ – T₁) + m_wc_w (T₃ – T₁)

                      (T2-T3)

 

Where CcCs and Cw are specific heat capacities of the calorimeter, the solid and water respectively.
The specific heat capacity of the material of the cube can therefore be calculated as;

1. A lagged copper calorimeter of mass 0.75 kg contains 0.9 kg of water at 20⁰C. A bolt of mass 0.8 kg is transferred from an oven at 400⁰C to the calorimeter and a steady temperature of 50⁰C is reached by the water after stirring. Given Specific heat capacity of copper is 400 Jkg-1K-1and that of water 4200 Jkg-1K-1. Determine:
2. Heat lost by the bolt

Let c be the specific heat capacity of the material of the bolt.

=0.8 x c x (400 – 50)

=280c

1. Heat absorbed by the water

=0.9 x 4200 x (50 – 20)

=113,400J

1. Heat absorbed by calorimeter

HC=mc∆T= 0.75 x 400 x (50 – 20)

=9000J

1. Specific heat capacity of the bolt

 

Solution

Heat lost by bolt = heat gained by calorimeter + heat gained by water

280c = 9000+ 113 400

280c = 122 400

c = 437 Jkg-1K-1

Specific heat capacity of bolt is 437 Jkg-1K-1.

 

 

Liquids

Using a solid of known specific heat the capacity and replacing the water in the calorimeter with liquid whose specific heat capacity is to be determined, the same procedure as in experiment 9.2 is repeated.

Calculations

Replacing the mass of water and the specific heat capacity of water with those of the liquid, ml and cl and making the same assumptions on heat losses to the surroundings, the specific heat capacity of the liquid can be calculated as below;

Heat lost by       =     heat gained by            + heat gained by liquid

The hot solid             calorimeter and stirrer     in the calorimeter

mscs (T2 – T3) = mccc (T3 – T1) + mlcl (T3 – T1)

mscc (T2 – T3) – mccc (T3 – T1) = mlct (83 – 81)

Hence, the specific heat capacity of the liquid can be calculated as;

    cl =   m_sc_s (T₂ – T₃) – m_cc_c (T₃ – T₁)

               ml(T3-T1)

Note:

The following precautions need to be taken to minimize heat losses to the surroundings:
(i) Use of a highly polished calorimeter.

(ii) Heavy lagging of the calorimeter.

(iii) Use of a lid of poor conduction.

Example 6

A block of iron of mass 1.25 kg at 120°C was transferred to an aluminium calorimeter of mass 0.3 kg containing a liquid of mass 0.6 kg at 25°C. The block and the calorimeter with its contents eventually reached a common temperature of 50 °C. Given the specific heat capacity of iron as 450 Jkg-1K-1 and that of aluminium 900 Jkg-1K-1, calculate ;

1. Heat lost by the iron

Hm=mc∆T =1.25 x 450 x 70

=39 375J

1. Heat absorbed by the liquid

Let c be the specific heat capacity of the liquid.

Hl=mc∆T=0.6 x c x 25

=15c

1. Heat absorbed by calorimeter

HC=mc∆T= 0.3 x 900 x 25

=6 750 J

1. Specific heat capacity of the liquid

Heat lost by iron block = heat gained by calorimeter + heat gained by liquid
1.25 x 450 x 70 = 0.3 x 900 x 25 + 0.6 x c x 25

39 375 = 6 750 + 15c

39 375 – 6 750 = 15c

c = 2 175 J kg-1K-1

 

1. A girl heats 5 kg of water to temperature of 800 When she adds m kg of water at 150 C the mixture attains temperature of 40⁰C. Determine the value of m. (ignore heat changes due to the container)           (3 marks)
2. Equal masses of water and paraffin with specific heat capacities C­Wand CP respectively are heated using identical sources of heat, for the same length of time. The final temperature θP of paraffin was found to be greater than final temperature  of water, Show that CW is greater than CP. (2 marks)

 

Electrical Method

Solids

EXPERIMENT; To determine the specific heat capacity of a metal by electrical method

Apparatus

Metal block with two holes, heater, thermometer, ammeter, voltmeter, stopwatch, lagging material, power source, rheostat, connecting wires, weighing balance.

Procedure and sample results

• Weigh the Mass of metal block = m
• Set up the apparatus as shown in figure
• Record the readings of the ammeter reading = I
• Record the readings of Voltmeter reading = V
• Start the stopwatch as you switch on the heater circuit.
• Record Time taken to heat the metal block = t
• Measure and record Initial temperature of the metal block = T1
• Measure and record Final temperature of the metal block = T2

Temperature change of the metal block = T2 – T1

Assuming no heat losses to the surroundings, the specific heat capacity of the material of the metal block is calculated as follows;

Electrical energy supplied by the electrical heater coil = heat gained by the metal block.
VIt = mc (T2 – T1)

where c is the specific heat capacity of the material of the block.
Therefore, specific heat capacity c is given by;

c =          VIt  

m(T2-T1)

Precautions

(i) The metal block must be highly polished and heavily lagged.

(ii)   The two holes should be filled with a light oil to improve thermal contact with the heater and thermometer.

Example 7

A metal cylinder of mass 0.5 kg is heated electrically. If the voltmeter reads 15 V, the ammeter 3.0 A and the temperature of the block rises from 20°C to 85 °C in 10 minutes, calculate the specific heat capacity of the metal cylinder.

Solution

Heat supplied by the heater = heat gained by the metal cylinder

VIt = mcΔT

15 x 3 x 10 x 60 = 0.5 x c x 65

c = 15 x 3 x 600

0.5 x 65

= 831 Jkg-1K-1

EXPERIMENT: To determine the specific heat capacity of a liquid by electrical method

Apparatus

Well-lagged calorimeter, stirrer, 12 V heating coil, liquid (water), ammeter, voltmeter, weighing
balance, thermometer, stopwatch, rheostat.

 

 

 

 

 

 

 

 

 

Fig. 9.4: Determination of specific heat capacity of a liquid by electrical method

Procedure

• Weigh the calorimeter with the stirrer. = m1
• Pour water into the calorimeter.
• Weigh the calorimeter with the water. = m2

Mass of water = m2 –  m1 = mw

• Place the Calorimeter in its insulating jacket.
• Measure the initial temperature of the water. = T1
• Insert the heating coil into the water in the calorimeter
• Switch on the heater current and simultaneously start timing.
• Record the ammeter = I
• Record voltmeter readings. = V
• Stir the water and after about five to ten minutes, switch off the heater current, but continue stirring and note the highest final temperature of the water.
• Record the duration of the heating. = t
• Record Final temperature of the water = T2

Assuming no heat is lost to the surroundings, the specific heat capacity of the water can be calculated as follows; ,

heat supplied by electric heater = heat gained by water + heat gained by calorimeter
Vlt = mwcw(T2 – T1) + m1cc(T2 – T1)

cw =  VIt – m(T₂-T₁)

            mw(T2-T1)

Where cw and cc are the specific heat capacities of water and material of the calorimeter respectively,

Example 8

In an experiment to determine the specific heat capacity of water, an electrical heater was used. If the voltmeter reading was 24 V and that of ammeter 2.0 A, calculate the specific heat capacity of water if the temperature of a mass of 1.5 kg of water in a 0.4 kg copper calorimeter rose by 6 ˚C after 13.5 minutes. (Specific heat capacity of copper is 400 Jkg+K:’)

Solution

Electrical energy supplied by heater = heat gained by calorimeter + heat gained by water 24 x 2 x 13.5 x 60 = 0.4 x 400 x 6 + 1.5 x c x 6, where c is specific heat capacity of water. 38 880 – 960 = 9c

c = 37920

9

= 4213 Jkg-1K-1

 

2002 Q19 P1

An immersion heater rated 90W is placed is a liquid of mass 2kg.  When the heater   is switched on for 15 minutes the temperature of the liquid rises from 20oC to 30oC Determine the specific heat capacity of the liquid. (Assume no heat losses)

 

Continuous Flow Method

EXPERIMENT 9.5: To determine the specific heat capacity of water by the continuous flow method

Apparatus

Constant head tank, electric heating coil, two thermometers, glass tube (surrounded by an evacuated glass jacket which prevents heat escape from the liquid by conduction or convection), stopwatch, rheostat.

 

 

Procedure

• Set up the apparatus as in figure 5.
• Adjust the water flow from the constant head tank until no water flows through the waste
  pipe (the water through the tube carrying the spiral heating coil has then attained a steady flow).
• Switch on the heater circuit and let the water flow continue until the thermometers at the inflow and outflow ends of the tube attain steady readings.
• Record the ammeter and voltmeter readings.
• Record the inflow and outflow temperatures of the water.
• Weigh an empty beaker and place it below the outflow pipe and simultaneously start timing.
• When an appreciable quantity of water has collected in the beaker, withdraw the flow pipe as you stop timing.
• Weigh the beaker with water and record the duration of heating.
• Repeat the procedure but with a different rate of flow of water.

Results and Calculations

First rate of flow of water

Ammeter reading = II

Voltmeter reading = VI

Inflow temperature of water = T1
Outflow temperature of water = T2
Mass of the empty beaker = ml
Mass of the beaker with water = m2
Time of collecting the water = t
Mass of water collected = m2 – ml = mc
Temperature difference = T2 – T1

Second rate of flow of water
Ammeter reading = I2

Voltmeter reading = V2

Mass of the beaker with water = m3
Mass of water collected = m3-m1= mn

Under steady conditions, none of the electrical energy supplied is used in heating the apparatus
and, therefore;

Electrical energy = heat energy absorbed    +        heat energy lost to

Supplied                   by collected water                  the surrounding

VIt = mw c(T2 – T1) + H

After the rate of flow is altered, temperature difference is the same and the heat lost in time t is again H.
:. V2I2t = mn c(T2 – T1) + H

Hence;

(V2 I2 – V1I1) t = (mn -mw) c(T2 – T1)

Advantages of the method over the other methods

• The presence of the vacuum prevents heat losses by convection or conduction.
• The steady temperature measured allows small temperature rises to be used and therefore suitable for determining the manner in which the specific heat capacity changes with temperature.
• Heat capacities of the apparatus not involved in the calculation.
• The method can be used for gases.

Exercise 9.1

1. A lady wanted to have a warm bath at 40°C. She had 5.0 kg of water in a basin at 85˚C. What mass of cold water at 25°C must she have added to the hot water to obtain her choice of bath? Neglect heat losses and take specific heat capacity of water as 4 200 Jkg-1K-1.
2. 2 kg of iron at 100°C is dropped into 0.09 kg of water at 26 °C inside a calorimeter of mass 0.15 kg and specific heat capacity 800 Jkg-1K-1. Find the final temperature of the water. (Specific heat capacity of iron is 460 J kg-1K-1and that of water 4200 Jkg-1K-1)
3. A copper calorimeter of mass 0.12 kg contains 0.1 kg of paraffin at 15°C. If 0.048 kg of aluminium at 100°C is transferred into the liquid and the final temperature of the mixture is 27°C, calculate the specific heat capacity of paraffin, neglecting heat losses. (Specific heat capacity of aluminium is 900 Jkg-1K-1and that of copper 400 Jkg-1K-1)

CHANGE OF STATE

• Heating a material generally leads to a rise in temperature. However, there are situations where no observable change in temperature is noted when a material is heated or cooled.
• When ice at -10⁰C is heated and temperature recorded at intervals as while stirring, the following observations are made

 

Observation

1. The thermometer records a temperature rise of the ice until it reaches O°C.
2. The temperature remains at 0 °C while ice changes to water at O°C.

• After all the ice has melted, temperature raises again,

 

Explanation

• When the ice at about -10 °C is heated, the heat energy is used in raising its temperature to °C.
• The energy given to the ice at 0 °C is used to change ice from solid to liquid state.

Conclusion

The heat supplied to ice at 0 °C does not change the temperature of the ice, but changes its state from solid to liquid (melting).

The heat absorbed as the ice melts is called latent heat. The term ‘latent’ means ‘hidden’. It is used thus because the ice at 0 °C is converted water at 0 °C without change in temperature.

Latent Heat of Fusion

Latent heat of fusion is defined as the heat required to change the state of a material from solid to liquid without temperature change. Conversely, as a liquid changes to solid state, latent heat of fusion is given out.

 

EXPERIMENT: To explore the change of state of naphthalene using the cooling curve
Apparatus

Naphthalene is heated to melt and then allowed to cool.

The temperature is recorded in intervals of time and a graph of temperature against time is plotted

 

                                      

Observation

During cooling, the temperature of liquid naphthalene falls from about 90°C to about 80 °C, where it remains constant for some time.

At this temperature, all the naphthalene gradually changes to solid, after which the temperature falls further to room temperature.

The graph of temperature against time is as shown in figure

Cooling curve for naphthalene

 

Explanation

• The portion OP represents the liquid naphthalene cooling.
• In PQ, the liquid naphthalene changes to solid without change in temperature.
• Point P is the freezing point, where the naphthalene is solidifying (and also the melting point).
• In portion QR, solid naphthalene cools to room temperature at R.
• During the period PQ, the liquid naphthalene is losing latent heat of fusion as it solidifies.

Specific Latent Heat of Fusion

the specific latent heat of fusion of a substance is defined as the quantity of heat energy required to change a unit mass of the substance from solid to liquid without change in temperature.

Q = mLf, where Lf is the specific latent heat of fusion. The SI unit of specific latent heat is Jkg-1.

Lf = Q/m

Note: A unit mass of a material changing from liquid to solid would give out heat energy equal to its specific latent heat of fusion.

 

 

 

EXPERIMENT: To determine the quantity of heat required to change unit mass of ice to water
Apparatus

Water, ice pieces, thermometer, calorimeter, stirrer.

 

 

Determining latent heat of fusion of ice
Procedure

• Pour water previously heated to about 5 °C above room temperature into a clean dry copper calorimeter of known mass.
• Pour water previously heated to about 5˚C above room temperature into a clean dry copper calorimeter of known mass
• Find the mass of the calorimeter with water.
• Record the temperature of the contents and add pieces of dry melting ice one at a time, each time stirring until the piece melts before adding the next.
• Continue the process until the temperature falls to about 5 °C below room temperature.
• Find the mass of the calorimeter with the mixture.

Note

1. Warming the water so that its temperature rises by a given value above room temperature then cooling it to a temperature which is the same value below room temperature balances the heat exchange between the calorimeter with its contents and the surrounding.
2. ‘Dry ice’ is one which has minimum water moisture on its surface. It is used so that any heat absorbed is utilized in changing of state from solid to liquid, but not in warming the water.

Results and Calculations

Mass of calorimeter and stirrer = m1
Mass of water and calorimeter =m2
Mass of calorimeter and mixture = m3
Temperature of water in calorimeter = T1
Final temperature of mixture = T2

Mass of water used = m2 – m1 = mw

Mass of ice melted = m3 – m2 = mm
Temperature change = T2 – T1

 

Heat lost   heat lost =     heat gained         +     heat gained      +              heat gained

(By warm water)          (By calorimeter)      (Ice at 0 °C to water at 0°C)    (Water at0 °C to final temp T2)

 

 

Let the quantity of heat required to melt a unit mass of ice at 0 °C to water at 0 °C be Lf
mw Cw (T1 – T2) + m2cc(T1 – T2) = mm Lf+ mw Cw (T2 -0)

where Cw and Cc are specific heat capacities of water and calorimeter material respectively.

Lf = m_w c_w (T₁ – T₂) + m_cc_c(T₁ – T₂) – m_w c_w T₂

                  mm

This is the quantity of heat energy required to melt unit mass of ice at constant temperature and is referred to as the specific latent heat of fusion of ice.

 

EXPERIMENT: To determine specific latent heat of ice by electrical method

Apparatus

Crushed ice, two filter funnels, two beakers, voltmeter, ammeter, rheostat, heater, two
thermometers, and stopwatch.

 

Put equal quantities of crushed ice into two identical filter funnels P and Q,

• Place an immersion heater connected to an ammeter, voltmeter and rheostat in P, making sure it is completely covered with ice.
• At the same time as you switch on the immersion heater, place dry empty beakers of known masses under P and Q.
• Note the reading of the ammeter and voltmeter (adjust the rheostat to keep them constant throughout the experiment).
• When a reasonable amount of water has collected in the beaker under P, note the time, remove the beakers and switch off the heater.
• Weigh the beakers and their contents.

Results and Calculations

Mass of beaker under P before experiment = m1
Mass of beaker under P after experiment =m2

Mass of ice melted in P during experiment = (m2 – m1) = mn
Mass of beaker under Q before experiment = m3

Mass of beaker under Q after experiment = m4

Mass of ice melted in Q during experiment = (m4 – m3) =mm
Reading of ammeter = I

Reading of voltmeter = V

Time during which heater is switched on = t seconds.

• The funnel Q and its contents is the control experiment.
• It enables the mass of ice melted due to the temperature of the room during the experiment to be obtained.
• It is reasonable to assume that the same mass will be melted in P.
• Thus, the mass of ice melted by the heater is mm- mn = m
  Then, heat energy supplied by the heater = heat energy gained by melting the ice
  VIt = mLf

Lf = VIt

         m

where Lf is the specific latent heat of fusion of ice.

The SI unit for specific latent heat of fusion is Jkg-1

Table gives values of specific latent heat of fusion of some common materials.

Material	Specific latent heat of Jus ion (x 105 Jkg-1}
Copper	4.0
Alwninium	3.9
Water (Ice)	3.34
Iron	2.7
Wax	1.8
Naphthalene	1.5
Solder	0.7
Lead	0.026
Mercury	0.013

Example 9

In an experiment to determine the specific latent heat of fusion of ice, 0.025 kg of dry ice at 0˚C is melted up in 0.20 kg of water at 21°C in a copper calorimeter of mass 0.25 kg. If the final temperature of the mixture falls to 11 ˚C, what is the specific latent heat of fusion of ice? (Take the specific heat capacity of water as 4200 Jkg-1K-1 and that of copper as 400 Jkg-1K-I).

Solution

Heat loss by +                                       heat lost by water   =    heat gained                           + heat gained

calorimeter                           by melting ice      by melted ice

 

0.25 x 400 x 10 + 0.2 x 4 200 x 10 = 0.025L + 0.025 x 4 200 x 11,

where L is the specific latent heat of fusion of ice.

1 000 + 8 400 = 0.025L + 1 155

0.025L == 9400 – 1 155

L == 8245

0.025

= 329 800 Jkg-1

= 3.298 x 105 Jkg-1

Hence the specific latent heat of fusion of ice is 3.298 x 1 ()5 Jkg-l.

Latent Heat of Vaporization

• The specific latent heat of vaporization (Lv) of a material is the quantity of heat required to change a unit mass of the material from liquid to vapour without change of temperature.
• The heat absorbed by a liquid as it change to vapour is the same heat energy given out by the vapour as it changes its state to liquid without change in temperature.

Q=mLv 

The SI unit of specific latent heat of vaporization is Jkg-1

 

EXPERIMENT: To determine the specific latent heat of vaporization of a liquid (water)
USING THE METHOD OF MIXTURES

Apparatus

Calorimeter, stirrer, water, thermometer, flask, delivery tube, heat source.

 

PROCEDURE

• Set up the apparatus as in figure above.
• Find the mass of the calorimeter when empty = mc
• Find the mass of the calorimeter when filled with water to the level shown. = m2
• Place the calorimeter with its contents in a felt jacket (lagging material).
• Record the initial temperature of water in the calorimeter. = T1
• Boil the water in the flask until steam starts issuing out freely through the delivery tube.
• Bring the free end of the delivery tube into the water in the calorimeter.
• Allow steam to bubble into the water while stirring until the temperature of the water rises by 20 ˚C above room temperature.
• Remove the delivery tube from the calorimeter and record the temperature of the water.
• Determine the mass of the calorimeter with condensed steam. = m3
• Determine the final temperature of condensed steam = T2

 

Results and Calculations

Mass of water = (m2 – mc) = mw

Mass of condensed steam =m3 – m2= ms
Temperature change = T2 – T1

 

• NB: When steam passes into the water, it first changes to water at 100 °C and then cools from 100 °C to final temperature of the mixture T2,
• Heat lost by condensing steam = msLv; where Lvis the specific latent heat of vaporization of steam.
• Heat lost by cooling water = mscw(100 – T2) where Cw is the specific heat capacity of water.
• Heat gained by calorimeter and stirrer = mccc(T2 – T1), where c is the specific heat capacity of the calorimeter.
• Heat gained by water = mwcw(T2- T1)

 

 

Therefore

 

Heat lost by steam + heat lost by water = heat gained by water   + heat gained by calorimeter

(condensation)          (100˚C-T2)    

msLv + mscw(100-T2)  =  mwcw(T2-T1) +  mccc(T2-T1)

 

Lv  =  m_wc_w(T₂-T₁) +  m_cc_c(T₂-T₁)- m_sc_w(100-T₂) 

                    ms

Note:

Errors due to heat loss to the surroundings can be minimized by first cooling the water in the calorimeter by a given value below room temperature and then passing the steam until the temperature rises above room temperature by the same value.

Example 10

Dry steam is passed into a well-lagged copper calorimeter of mass 0.25 kg containing 0.50 kg of water and 0.02 kg of ice at 0 ⁰C. The mixture is well stirred and the steam supply cut off when the temperature of the calorimeter and its contents reaches 25°C. Neglecting heat losses, if 25g of steam is found to have condensed to water and Specific heat capacity of copper is 400 Jkg-1K-1and latent heat of fusion of water is 3.36 x105 Jkg-1) find:

1. Heat lost by the steam

H=m Lv

=0.025Lv

1. Heat lost by hot water

H=mc∆T

=0.025 x 4 200 (100 ⁰C – 25°C)

=7 875J

1. Heat gained by calorimeter

H=mc∆T

=0.25 x 400 x 25

=2500J

1. Heat gained by ice

H=m Lf

=0.02 x 336000

=6 720J

1. Heat gained by cold water

H=mc∆T

=(0.5 + 0.02) 4 200 x 25

=54600J

1. the specific latent heat of vaporization of water

 

Solution

heat lost by steam+ heat lost by water =  heat gained    +  heat gained by      + heat gained by

(condensation)        (100 ⁰C – 25 ⁰C)      by ice (fusion)    water (0 ⁰C – 25 ⁰C)     calorimeter

 

Let L be the specific latent heat of vaporization of water + heat gained by

0.025L + 7 875 = 6 720 + 54 600 + 2 500 0.025L = 55 945
L= 55945
     0.025

= 2237800 Jkg-1.

Specific latent heat of vaporization of water is 2.238 x 106 Jkg-1.

 

Determining specific latent of vaporization of water using ELECTRICAL METHOD

Beaker, flask, heater coil, condenser, water, stop watch, voltmeter, rheostat, ammeter power source.

 

                                       

Procedure

• Set up the apparatus as shown in figure 9.13.
• Switch on the heater and maintain a steady current using the variable resistor.
• Allow the heating to continue until the system reaches a steady state, where condensed water issues down the tube T at a constant rate.
• Weigh the beaker and place it under the tube to collect the condensed water and simultaneously start timing.
• When a measurable quantity of water has collected in the beaker, remove the beaker as you stop the watch.
• Record the time taken and weigh the beaker with condensed water.

Results and Calculations

Ammeter reading = I

Voltmeter reading = V

Mass of empty beaker = mB

Mass of beaker with condensed water = m2
Time taken to collect the condensed water = t
Mass of condensed water = m2 – mB= ms

Assuming that all the heat given by the heater coil is used in vaporizing the water and all the steam is condensed, the specific latent heat of vaporization of water is calculated as follows;
heat supplied by the heater coil = heat used to vaporize the water.

VIt = mcLv 

Lv = ( VIt ) ,

            mc   where Lv is the specific latent of vaporization.

 

 

Table 9.3 gives values of specific latent heat of vaporization for some common materials.

Material	Specific latent of vaporization
	( x 105 Jkg-1)
Water	22.6
Alcohol	8.6
Ethanol	8.5
Petrol	6.3
Benzene	4.0
Ether	3.5
Turpentine	2.7

 

Example 11

Calculate the quantity of heat required to change 0.50 kg of ice at -10°C completely into steam at 100 °C Take:

specific heat capacity of ice = 2 100 Jkg-1K-1

specific heat capacity of water = 4 200 Jkg-1K

specific latent heat of fusion of ice = 3.36 x 1()5 Jkg-1
specific latent heat of vaporisation of steam 2.26 x 106 Jkg-1

Solution

Quantity of heat gained by ice to raise its temperature to 0˚C
mcΔT = 0.50 x 2100 x 10

= 10500J

Quantity of heat required to change the ice into water;
mL= 0.50 x 336 000

= 168000 J

. Quantity of heat required to raise temperature of water to 100 DC;
mcΔT = 0.50 x 4 200 x 100

= 210 OOOJ

Quantity of heat required to vaporize water;
mLv = 0.5 x 2 260 000

= 1130000J

Total heat energy = 10 500 + 168000+ 210 000 + 1 130000
= 1518500 J

 

Factors Affecting Melting and Boiling Points

1. pressure on melting point

Apparatus

Block of ice, thin copper wire, two heavy weights, and wooden support.

 

Procedure

• Attach two heavy weights to the ends of a thin copper wire.
• Pass the string over a large block of ice, as shown above

Observation

The wire gradually cuts its way through the ice block, but leaves it as one piece.

Explanation

• Because of the hanging weights, the wire exerts pressure on the ice beneath it and therefore
  makes it melt at a temperature lower than its melting point.
• Once the ice has melted, the water formed flows over the wire and immediately solidifies since it is no longer under pressure.
• As it solidifies, the latent heat of fusion is released and is conducted by the copper wire to melt the ice below the copper wire.
• The process continues until the wire cuts through leaving the block intact.

Conclusion

Application of pressure on ice lowers the melting point.
Note:

• The thermal conductivity of the wire used in this experiment plays a crucial role.
• If an iron wire with lower conductivity is used, it will cut through, but much more slowly.
• Cotton string, which is a poor conductor, will not cut through the ice block at all.

The process of refreezing is known as regelation.

Applications of the Effects of Pressure on Melting Point of Ice
Ice Skating

• The weight of an ice-skater acts on the thin blades of the skates.
• The high pressure exerted by the thin blades melts the ice underneath, forming a thin film of water over which the skater slides.

Joining Ice Cubes under Pressure

• Two ice cubes can be joined together by pressing them hard against each other.
• The increased pressure lowers the melting point of the ice at the points of contact. With the pressure lowered, the water recondenses and the two cubes join together.

 

2. Impurities

• The presence of impurities lowers the melting point of a substance.
• An application of this is where salt is spread to prevent freezing on roads and paths during winter. Similarly, a freezing mixture can be made by mixing ice with salt.

 

 

BOILING

1. a)Increasing the pressure

 

 

 

 

 

 

 

 

 

 

 

 

 

Procedure

• When the water begins boiling and steam issues from the rubber tube steadily, squeeze
  the rubber tube momentarily and observe the thermometer shows a rise in temperature and the boiling reduces.

Explanation

• Closing the rubber tube causes an increase in vapour pressure within the flask.
• This makes it more difficult for molecules from the surface of the liquid to escape, raising the boiling point of the liquid.

Conclusion

• Increase in pressure increases boiling point of a liquid.
• An application of this concept is the pressure cooker. It has tight-fitting lid which prevents free escape of steam, thus making pressure inside it build up.
• The boiling point is increased to a higher temperature, enabling food to cook more quickly.

 

b)Reducing the pressure

Procedure

• Heat water in a round-bottomed flask fitted with a thermometer and a tube.
• Stop heating, close the clip and turn the flask upside down.
• Run cold water over the flask,

 Observation

• The water stops boiling when the heating is stopped. When cold water is then run over the flask, the water inside begins boiling again, even though its temperature is below boiling point.

Explanation

• The cold water condenses the steam and therefore reduces vapour pressure inside the flask.
  This lowers the boiling point of the liquid.

Conclusion

• Decrease in pressure lowers the boiling point of a liquid.

2. the effect of impurities on boiling point

 

 

 

 

 

Procedure

• Heat some distilled water in a beaker and note the boiling point.
• Make a strong salt solution and repeat the experiment .
• Record the new boiling point.

Observation

• The boiling point of the salt solution is higher than that of the distilled water.

Conclusion

• The presence of impurities in a liquid raises its boiling point.

 

Evaporation

• Molecules in a liquid are in a continuous random motion with varying kinetic energy.
• evaporation This process in which molecules at the surface of the liquid acquire sufficient kinetic energy to overcome the attractive force from the neighboring molecules in the liquid, and thus escape
• it occurs at all temperatures, even before boiling point.

 

1. Effects of Evaporation

• Pour some methylated spirit on the back of your hand. The hands feel cold as the spirit evaporates from the skin. The evaporating methylated spirit extracts latent heat from the skin, making it feel cold.

(ii)           In a fume chamber, pour some ether into a test tube. Bubble air through the ether using a long rubber tubing.

Frost forms on the outside surface of the tube.

 

• The evaporating ether draws latent heat of vaporization from the liquid ether, the test tube and the surrounding space.
• The tube therefore cools so that frost forms around it. Bubbling increases the surface area of ether exposed to air.

 

(iii) Place a beaker on film of water on a wooden block. Pour some ether into the beaker and blow air through the ether using a foot pump.

The ether quickly evaporates and after sometime, it is found that the beaker is stuck to the wooden block, a thin layer of ice having formed between them. This shows that evaporation causes cooling.

 

Factors affecting Rate of Evaporation

Temperature

Increasing the temperature of a liquid makes its molecules on its surface move faster. This
makes it easier for more of them to escape, enhancing evaporation. It takes shorter time for
cloths to dry on a hot day.

Surface Area

Increasing the area of the liquid surface gives the faster molecules greater chance of escaping.
A wet bed-sheet dries faster when spread out than when folded.

Draught

Passing air over the liquid surface sweeps away the escaping vapour molecules. This clears the way for more escaping molecules to enter the space. This is why wet clothes dry faster on a windy day.

Humidity

Humidity is the concentration of water vapour in the atmosphere. When humidity is high, there are more vapour molecules in the space above the liquid surface. This makes it more difficult for the water molecules to leave the surface. Wet clothes take longer time to dry up on a humid day.

Comparison between boiling and evaporation

 

Evaporation	Boiling
Takes place at all temperatures	Takes place at a fixed temperature.
Takes place on the surface of the liquid.	Takes place throughout the liquid, with
No bubbles are formed.	bubbles of steam forming all over.
Decreasing the atmospheric pressure	Decreasing atmospheric pressure
increases the rate of evaporation.	lowers the boiling point.

Application of Cooling by Evaporation
Sweating

When sweat evaporates, it draws latent heat from the skin, producing a cooling effect. Animals have different mechanisms of cooling their bodies. A dog exposes its tongue when it is hot, while the muzzle of a cow gets more wet when it is hot.

Cooling of Water in a Porous Pot

A porous pot has tiny pores, through which water slowly seeps out. When this water evaporates, it cools the pot and its contents.

The Refrigerator

The effect of cooling caused by evaporation is made use of in the refrigerator.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

• In the upper coil, the volatile liquid (Freon) takes latent heat from the air around and evaporates, causing cooling in cabinet.
• The vapour is removed by the pump into the lower coil outside the cabinet, where it is compressed and changes back to liquid form.
• During this process, heat is given out and conducted by the copper fins to the surrounding air.
• The liquid goes back to the copper coil and the cycle is repeated.
• There is also a thermostat which controls the rate of evaporation and hence the temperature inside the refrigerator.

 

Figure 11 shows the features of a domestic refrigerator. A volatile liquid circulates through the capillary tubes under the action of the compression pump.

 

 

 

 

 

 

 

 

 

(i) State the reason for using a volatile liquid                                                            (1 mark)

So that it vaporizes readily/ easily                                              

(ii) Explain how the volatile liquid is made to vaporize in the cooling compartment and to condense in the cooling fins                                 (2 marks)

In the freezing compartment the pressure in the volatile liquid is lowered suddenly by increasing the diameter of the tube causing vaporization in the cooling fins, the  pressure  is increased by the compression pump and heat lost to the outside causing condensation.

Acquires heat of the surrounding causing the liquid to vaporize

     (iii) Explain how cooling takes place in the refrigerator                                             (3 marks)

When the volatile liquid evaporates, it takes away heat of vaporization to form the freezing compartment, reducing the temperature of the latter. This heat is carried away and disputed at the cooling fins where the vapour is compressed to condensation giving up heat of vaporization

     (iv)    What is the purpose of the double wall?                                              (1 mark)

Reduces rate of heat transfer to or from outside ( insulates)

                           Reduces / minimizes, rate of conduction/ conversion of heat transfer                        

 

 

 

 

 

 

Revision Exercise 9

(Take specific heat capacity of water as 4 200 Jkg+K:’)

 

1. Calculate the specific heat capacity of paraffin if 22 000 J of heat is required to raise the
   temperature of 1.5 kg of paraffin from 23°C to 30 °C.

 

1. A block of metal of mass 5 kg is heated to 110°C and then dropped into 1.5 kg of water.

 

The final temperature is found to be 50°C. What was the initial temperature of the water?
(The specific heat capacity of the metal is 460 Jkg-1K-I)

 

3. Water drops from a waterfall 84 m high. The temperature of the water at the bottom is
   found to be 26.2 0c. Calculate its temperature at the top.