MEASUREMENT OF HEAT ENERGY

OBJECTIVES
Students should be able to
i. Heat energy, heat capacity and specific heat capacity
ii. Explain how to determine the specific heat capacity of a solid by mixture and
electrical methods
iii. State the factors that affect heat enenrgy of a substance
iv. Solve problems relating to heat capacity

SPECIFIC HEAT CAPACITY AND HEAT CAPACITY

Heat is energy transferred as a result of temperature difference between two points of a
body. It is also the energy transmitted from a hot object to a cold object. It is the total energy of
a body (sum of kinetic and potential energy of a body). The amount of heat energy in a body is
measured by its mass if the temperature is constant. This is called thermal energy or the
quantity of heat and is measured in joule (J).

Heat energy and factors affecting it

Heat is the total energy of a substance. The quantity of heat in a body depends on:
(1) mass of the substance: The heat content of a substance is proportional to its mass if the
temperature is kept constant. A bucket of boiling water contains more heat than a cup of
boiling water. (Q α m)
(2) rise in temperature of the substance (Δθ): the higher the temperature of a substance, the
greater the heat content of the substance. A cup of water at a temperature of 100°C contains
more heat energy than the same cup of water at 30°C.

(3) the nature of the substance: If the same mass of copper and water are heated to the same
temperature, the heat content of water is higher than the heat content of copper. This shows
that the quantity of heat in a body depends on the nature of the substance.

Specific heat capacity (c)

Heat energy possessed by a body is directly proportional to the product of its mass and the
change in temperature.

The constant c is the specific heat capacity of the substance/body; it depends on the nature of
the substance.
Specific heat capacity of a substance is the quantity of heat energy needed to change the
temperature of a unit mass (1 kg) of a substance by 1 Kelvin.

The SI unit of specific heat capacity c is joule per kilogram per Kelvin (J kg-1 K-1). It is also
measured in joule per gram per Kelvin (J g-1 K-1).
Heat capacity (C)
Heat capacity is the quantity of heat needed to change the temperature of any mass of a
substance by 1K.
Heat capacity = Mass of substance × specific heat capacity of the substance
C = mc
Heat capacity of a substance depends on its mass. Heat energy contained in a given mass of a
substance is high if the heat capacity is high. For the same mass of copper and water, the rise in
temperature is higher for copper than water when they are given the same amount of heat
energy. This is because copper needs less heat energy to increase its temperature. The quantity
of heat absorbed or given out by a substance is equal to:
Q = mcΔ θ Or Q = CΔ θ
Worked examples
1. How much heat is needed to increase the temperature of 150g of water from 30°C to 50°C,
if the specific heat capacity of water is 4200 J kg-1 K-1?
Solution
Mass of water = 150 g = 0.15 kg
Increase in temperature Δθ = 50 – 30 = 20°C
Specific heat capacity c = 4200 J Kg-1 K-1
Q = mcΔθ
Q = 0.15 ×4200 ×20 = 12600 J
2. A copper sphere of mass 200 g is cooled from an initial temperature ϴ to 30°C. Calculate
the value of the initial temperature if the heat energy lost by the sphere is 72000 J. (Specific
heat capacity of copper is 400 J Kg-1 K-1)
Solution
Mass of copper = 200 g = 0.2 kg
Change in temperature Δ θ= (θ–30)
Specific heat capacity of copper c = 400 J kg K-1
Q = mcΔ θ
72000 = 0.20 × 400 × (θ- 30)
72000 = 80(θ-30) 800 = 72000 + 2400
θ = 930°C.

Heat exchange
Heat energy is exchanged anytime hot and cold substances mix. The hot substance loses heat
energy while the cold substance gains heat energy. Exchange of heat energy will continue until
the hot and the cold substances reach an equilibrium temperature. This is the maximum steady
temperature of the mixture. Energy is conserved, if the loss of heat to the surrounding is
minimized. The important assumption made in all heat exchange calculation is that:
Total heat energy lost by the hot substance is equal to the total heat energy gained by the
cold substance.
To carry out experiments on heat exchange in the laboratory, calorimeters are used.
Calorimeters are vessels made from metals like copper or aluminium. It is designed to ensure
that heat loss to the surroundings is minimized. Exchange of heat energy with the surroundings
is possible by radiation, conduction, convection and evaporation. Calorimeters are designed to
reduce heat lost to the environment by any of this mode of heat transfer.
• Heat loss by radiation is reduced by polishing or silvering the inner and outer calorimeter.
• Heat loss by conduction is reduced by surrounding the inner calorimeter with a bad
conductor like wool or cotton. This is called lagging.
• Heat lost by convection and evaporation is reduced by covering the calorimeter with a lid.

Figure 1: Calorimeter
Worked examples
3. An aluminium solid at 100°C is dropped into a copper calorimeter of mass 60g containing
80 g of water at 20°C. Calculate the mass of the aluminium solid if the steady maximum
temperature of the mixture is 30°C. {Specific heat capacity of copper, aluminium and water are
400, 900 and 4200 J Kg-1 K-1 respectively}
Solution

2. (a) Explain the meaning of the statement, “the specific heat capacity of water is 4200J kg -1”;
(b) 500 g of aluminium block at a temperature of 650°C is dropped into a copper calorimeter of
mass 100 g containing 800 g of water at 30°C. Calculate the maximum temperature of the
mixture. {Specific heat capacity of water, copper and aluminium are 4200 J kg-1 K-1, 400 J kg-1 K-1
and 900 J kg-1 K-1 respectively.}
Solution
(a) The statement means that 4200 J of heat energy is needed to increase the temperature of 1
kg of water by 1K.
(b) Heat lost by aluminium Qa =macaΔθa
Heat gained by calorimeter Qc = mcccΔθc
Heat gained by water Qw = mwcw Δθ
Heat lost by hot aluminium = Heat gained by copper calorimeter and water
maca Δ θa = mccc Δ θc+ mwcwΔ θw
0.5×900× (650 -θ) = 0.1×400×(θ- 30) + 0.8×4200 (θ-30)
450(650 – θ) = 40(θ- 30) + 3360(θ – 30) 292500 – 4500 = 400- 1200 + 33600- 100800
38500=39450θ
θ= 102.5°C
3. When 342000 J of heat energy is added to a material of mass 15 kg the temperature rises
from 30°C to 90°C. Calculate the heat capacity of the material.
Solution

Measurement of specific heat capacity of solid by method of mixture

Apparatus: A thin walled calorimeter, stirrer, thermometer, metal solid, tripod stand, Bunsen
burner, a beaker with water, beam balance and thread.
Method:
(i) Use the beam balance to carry out the following procedures:
• Weigh the solid provided and record its mass (ms);
• Clean the calorimeter, put the stirrer inside, weigh and record the mass of the empty
calorimeter and stirrer (mc);
• Fill the calorimeter with water enough to cover the solid, weigh again and record the mass
of calorimeter and water (m). Find the mass of water by evaluating mw = (m – ms). Measure and
record the initial temperature (θ1) of water and calorimeter.

Figure 2: Determination of specific heat capacity of a solid by method of mixture

(ii) Use the thread to suspend the solid in a beaker of water, allow standing for few minutes
until the solid and water attain an equilibrium temperature (θ1). Record this temperature θ1 as
the initial temperature of solid.
(iii) Put the beaker and its content on the tripod stand and perform the following procedures:
• Heat the beaker and its content gently till the water boils for several minutes to allow the
solid to reach the same temperature as the boiling water;
• Measure the temperature of the boiling water and record it as the temperature of the solid
(θ2=100°C);
• Transfer the solid quickly to the calorimeter after shaking off water droplets from the solid;
• Stir the water gently to maintain uniform temperature and record the highest steady
temperature of the mixture (θ).
Calculations:
Mass of water mw= (m – mc)
Change in temperature of solid Δθs = (100 – θ2)
Change in temperature of water Δθw = (θ2 – θ1) Change in temperature of calorimeter Δθw= (θ2-
θ 1)
Heat lost by hot solid = mscs(100 – θ2)
Heat gained by calorimeter = mccc(θ2 – θ1)
Heat gained by water = mwcw(θ2 – θ1)
Heat lost by solid = Heat gained by calorimeter and water
mscs(100 – θ) = mwcw(θ2 – θ1)+ mccc(θ2 – θ1)
Precautions
1. Transfer the hot solid quickly from the boiling water to the calorimeter and the lid covered
immediately to reduce loss of heat to the surrounding.
2. Drop the hot solid gently into the calorimeter to avoid splashing of water.
3. Stir the calorimeter and its content continuously to keep the temperature constant.
4. Lagged the calorimeter well to reduce heat exchange with the surroundings.

Measurement of specific heat capacity of liquid by method of mixture

Apparatus: A thin walled calorimeter, stirrer, thermometer, liquid (water), tripod stand, Bunsen
burner, a beaker with water, beam balance and thread.
Method:
(i) Clean the calorimeter, put the stirrer inside, weigh and record the mass of the empty
calorimeter and stirrer (mc);
(ii) Put the calorimeter inside a lagging material and measure its temperature θc;
(iii) Heat some quantity of water in a beaker, stir continuously to keep the temperature
uniform, measure and record the temperature of the hot water θ w;
(iv) Pour the hot water into the calorimeter until it is half filled, stir and measure the final
steady temperature (θ) of the mixture.
(v) Remove the calorimeter from its lagging; allow cooling to room temperature, measure and
record the mass of the calorimeter and water with beam balance.
Calculations:
Fall in temperature of water Δθw = θw – θ
Rise in temperature of calorimeter Δθc = θc – θ
Heat lost by hot water = mw cw (θw – θ)
Heat gained by calorimeter = mccc(θc – θ)
Heat lost by hot water = Heat gained by calorimeter
mw cw (θw – θ) = mccc(θc – θ)

Electrical method of measuring specific heat capacity of a liquid

Electrical energy
Electrical energy is converted to heat energy when electric current flows through a heating coil.
The quantity of heat produced is given by:
Q = IVt {I = current, V = voltage and t = time}
The heat produced is used to warm the water and increase its temperature.
Apparatus: Lagged calorimeter, heating coil, liquid (water), ammeter, voltmeter, rheostat, key,
battery, stop clock and thermometer.
Method
Figure 3: Measuring specific heat capacity of liquid
(i) Clean the calorimeter put the stirrer inside, weigh and record the mass of the empty
calorimeter and stirrer (mc);
(ii) Pour the liquid into the calorimeter until it is half filled, stir and measure the temperature
(θ1) of the liquid and calorimeter.
(iii) Switch on the current by closing the key and at the same time start the stop clock. Measure
and record the current through the ammeter and the voltage indicated by the voltmeter.
(iv) The liquid is stirred continuously to keep the temperature constant. The current is switched
off after t seconds and the final steady temperature of the liquid and calorimeter measured
(θ2).
Calculations
Heat given out by the heating coil = IVt
Heat gained by liquid = mlclΔ θl
Heat gained by calorimeter = mcccΔ θc
Heat given out by heating coil = Heat gained by calorimeter and water
IVt = mlclΔ θl+ mcccΔ θc
Precautions
1. To compensate for heat exchange with the surroundings, the liquid is first cooled to about
5°C below room temperature and heating continued until the temperature rises to about 5°C
above the room temperature.
2. The current flowing through the coil is kept constant through out the heating.
3. The liquid is stirred continuously to ensure uniform temperature throughout the volume of
the liquid.
4. The calorimeter should be well lagged.

Electrical method of measuring specific heat capacity of a solid

Apparatus: Lagged calorimeter, heating coil, metal block, ammeter, voltmeter, rheostat, key,
battery, stop clock and thermometer.
Method:

Figure 4: Electrical method of measuring specific heat capacity of a solid

The specific heat capacity of a metal block can be determined electrically as shown in Figure
8.4. Two holes are drilled on the block and lubricated with oil to allow the heater and the
thermometer make good thermal contact with the block. Exchange of heat with the
surroundings is reduced by lagging.
The current is switched on and kept constant throughout the heating process and at the same
time the stop clock is started. The current is allowed to flow for some minutes before it is
switched off before the readings of the ammeter and voltmeter are taken.
Calculations:
Heat given out by the heating coil = IVt
Heat gained by solid = mscsΔ θs
Heat given out by heating coil = Heat gained by the solid block
IVt = mscsΔ θs
Worked examples
1. An electric kettle of mass 0.5 kg rated 1 kW contains liquid of mass 1.5 kg. Calculate the
specific heat capacity of the liquid if it is heated from 30°C to 80°C in 5minutes. {Specific heat
capacity of the material of the kettle is 900 J kg-1 k-1}.
Solution
Power P of the kettle’s heater = IV = 1 000 W
Heat given out by the heater of the kettle: Q1 = Pt =1 000 × 5 × 60 = 300 000 J
Heat absorbed by the material of the kettle: Q2 = mkckΔθk = 0.5×900×50 = 22 500 J
Heat gained by liquid: Q3 = mlclΔθ1 = 1.5×c1× 50 = 75 c1
Heat given out by the heating coil = Heat gained by the liquid and kettle

2. Calculate the time it takes 1000 W heater to increase the temperature of 30 kg of copper
block from 25°C to 125°C, if the specific heat capacity of copper is 400 J kg-1 K-1.
Solution
Power P of the heater = IV = 1000 W
Heat given out by the heater Ql= Pt = 1000 t.
Heat absorbed by copper block Q2 = mcccΔθc = 30×400×100 = 1200000 J
Pt = mcccΔθc
1000t= 1200000
t = 1200 seconds = 20 minutes

Conversion of potential energy to heat energy

The energy of water at the top of waterfalls and dams is potential energy. The potential energy
of the water is transformed to kinetic energy as the water falls. At the bottom of the waterfall,
the kinetic energy is changed to heat energy, therefore the temperature of water at the base of
the waterfall is higher than the temperature at the top. The conversion from potential energy
to heat energy can be used to find the specific heat capacity of a substance.
Calculations
Potential energy at the top = mgh
Gain in heat energy at the base = mscsΔθs
Loss in potential energy at the top = Gain in heat energy at the base
mgh = mscsΔθs
Worked example
A brass sphere of mass 50 kg and specific heat capacity 380 J kg-1 K-1 dropped from a height of
500 m. What is the rise in temperature of the sphere if all the potential energy at the top is
transformed to heat energy?
Solution
Potential energy of the sphere at the top Ep = mgh = 50 × 10 × 500 = 250000 J
Gain in heat energy Q = mscsΔθs =50×380×Δ θs
Loss of potential energy = Gain in heat energy
mgh = mscsΔθs
50 × 10×500 = 50×380×Δ θs
250000= 19000Δ θs
Δ θs = 13.2°C
Significance of specific heat capacity
(i) Land and sea breeze
The same mass of water (specific heat capacity = 4200 J kg-1 K-1) needs more heat to increase its
temperature by 1°C compared to the soil (specific heat capacity = 800 J kg-1 K-1). During the day,
the temperature of the land rises more than that of water causing a current of hot air to move
from the land to the sea while cool breeze blows from the sea to the land. Land breeze (cool
air) blows from the land to the sea in the night.
(ii) Choice of liquid coolant
Liquid with high specific heat capacity like water are used as coolants in car radiators and heat
exchangers. This is because more heat energy is required to increase their temperature