**MEASUREMENT OF HEAT ENERGY**

Heat energy is the energy that is transferred from a hotter body to a cooler one as a result of their temperature difference. Heat can be used to do some work and could also be transformed to another one. It is measured in Joules, Kilojoules, etc.

**SPECIFIC HEAT CAPACITY OF A SUBSTANCE**

This is defined as the quantity of heat that is required to raise a unit mass or 1kg mass of a substance by 1K. it is denoted by “c” and measured in J/Kg/K. It is dependent on the nature of material the body is made of.

Mathematically,

c = Q/Δθ, where Q = Quantity of heat energy, m = mass of the body, Δθ = Temperature difference of the body.

**THERMAL/HEAT CAPACITY**

This is defined as the quantity of heat energy that is required to raise the temperature of a body by 1K. it is denoted by C_{p}. It is measured in J/K.

Mathematically,

C_{p} = mc = Q/ Δθ.

Example: A copper rod with heat capacity 585J/K is heated until the temperature changes from 35K to 80K. Calculate the quantity of heat supplied to the rod if the specific heat capacity of copper is 390J/Kg/K. Find the mass of the copper rod.

Solution

Given: Δθ = 80 – 35 = 45K, C_{p} = 585J/K, c_{c} = 390J/Kg/K

Required: i. Q =?; ii. m =?

Analysis:

- Q = C
_{p}x Δθ

= 585 x 45

= 26, 325J

- C
_{p}= mc

m = C_{p}/c

m = 585/390

m = 1.5Kg

**EXPERIMENT TO MEASURE THE SPECIFIC HEAT CAPACITY OF A SOLID BY THE ELECTRICAL METHOD**

From the set up above,

Let mass of block = m (Kg)

Let value of steady current = I (A)

Let value of potential difference across heater = V (Volts)

Let temperature rise of the block = θ (^{0}c)

Let time of flow of current = t (s)

Let S.H.C. of block = c (J/Kg/K)

Let heat supplied by the heater = H = IVt

Let heat required to raise the temperature of the block by θ^{0}c = Q

Assuming no heat lost to the surrounding,

Heat supplied by the heater = Heat required to raise the temp. of the block, i.e.

H = Q

IVt = mcθ

C = IVt/ mθ

Note: In the experiment above,

- A little oil is put in each hole to establish a good thermal contact with the block.
- The metal block is put inside a jacket lag to reduce heat loss.
- The metal block is weighed after boring two holes in the block.

PRECAUTIONS

- Lagging of metal is necessary to prevent heat lost to the surroundings.
- Accurate reading of the current, voltage and time must be taken.

Example: An electric heater rated 15V, 40W fitted into a metal block supplies heat to the block of mass 1.5Kg and S.H.C. of 460J/Kg/K. Calculate the temperature rise in the block if the current flows for 10mins.

Solution

Given: V = 15V, P = 40W, t = 10mins = 10 x 60 = 600s, c = 460J/Kg/K, m = 1.5Kg

Required: θ =?

Analysis:

c = IVt/ mθ

But, P = IV

θ = Pt/mc

θ = 40 x 600/ (1.5 x 460)

θ = 34.78^{0}c.

**EXPERIMENT TO MEASURE SPECIFIC HEAT CAPACITY OF A LIQUID BY ELECTRICAL METHOD**

Procedure

- A plastic container is first weighed and re-weighed when about 2/3
^{rd}full of liquid is put inside it. - Ammeter, Voltmeter, Heating coil are connected as shown in the circuit above.
- A plastic stirrer and a thermometer are also fitted into the container through holes in the wooden lid.
- Take the initial temperature of the liquid and record it.
- Switch on the current by connecting to the cell as shown
- Adjust the rheostat to give suitable steady current for about 3mins.
- Stir the liquid gently while heating to ensure uniform distribution of heat.
- Read the final temperature immediately after switching off the current.

Hence, from the experiment above, the following could be deduced to determine c;

Mass of liquid = m (Kg)

Initial temperature of liquid = θ_{1} (^{0}c)

Mass of the container = m_{c} (Kg)

Final temperature of liquid = θ_{2} (^{0}c)

Current flowing in circuit = I (A)

Voltage across heating coil = V (Volts)

Time of flow of current = t (s)

Specific heat capacity of the container = c_{c} (J/Kg/K)

At Equilibrium,

Heat supplied by the heater = Heat gained by the liquid and the container

IVt = mc (θ_{2 }– θ_{1}) + m_{c} c_{c} (θ_{2 }– θ_{1})

c = (IVt – m_{c} c_{c} (θ_{2 }– θ_{1}))/ m(θ_{2 }– θ_{1})

Note: (i.) In order to reduce heat loss to the surrounding, the water is first cooled with ice to about 10^{0}c below room temperature before the start of the experiment, and heating the water until its final temperature is about 10^{0}c above the room temperature.

(ii.) Errors can also arise when the heat capacity of the plastic container is neglected in the calculation.

PRECAUTIONS

- Lagging of copper calorimeter is necessary.
- Thorough stirring is important to ensure uniform temperature before the final reading is taken.
- The thermometer must be kept vertical before taking the reading to avoid error due to parallax.
- Avoid error due to parallax while taking the reading of the voltmeter and ammeter.

**WAYS BY WHICH HEAT CAN BE LOST TO THE SURROUNDING USING CALORIMETER**

- Conduction
- Convection
- Radiation
- Evaporation

**METHODS OF REDUCING HEAT LOSS**

- Losses due to conduction can be reduced by surrounding the calorimeter with a poor conductor e.g. cotton wool through the process called “Lagging”.
- Losses due to convection and evaporation can be reduced by covering the calorimeter with a lid containing holes to take a thermometer and a stirrer.
- Losses due to radiation can be reduced by polishing the outside of the calorimeter.

Calorimeter: This is an apparatus used in experiment that involves the exchange of heat between two bodies at different temperatures. It is usually made of copper and aluminium.

**EXPERIMENT TO DETERMINE THE SPECIFIC HEAT CAPACITY OF A SOLID BY METHOD OF MIXTURES**

PROCEDURE:

- A copper/ brass block is weighed, tied to a string and left for some minutes in a beaker of boiling water.
- Weigh the calorimeter together with the stirrer along and re-weigh when it is about 2/3
^{rd}full of water. - Read the initial temperature of the water with thermometer.
- Allow the block to stay in the boiling water for about 15mins.
- Transfer quickly into the water in the calorimeter and cover the calorimeter with the lid.
- Stir the water gently to ensure a uniform distribution of heat.
- Read and record the highest steady temperature of the mixtures.

Hence, from the experiment above, the following could be deduced;

Mass of copper block = m_{1} (Kg)

Mass of calorimeter + stirrer = m_{2} (Kg)

Mass of calorimeter + stirrer + water = m_{3} (Kg)

Initial temperature of water + calorimeter = θ_{1}

Final temperature of the mixture = θ_{2}

Temperature of boiling water = 100^{0}c

S.H.C of solid = c_{1}

S.H.C of water = 4200 J/Kg/K

S.H.C of calorimeter material = c_{2}

Neglecting any heat losses to the surroundings;

Heat lost by copper block = Heat gained by water + Heat gained by calorimeter and stirrer

m_{1} c_{1 }(100 – θ_{2}) = (m_{3} – m_{2}) x 4200 x (θ_{2} – θ_{1}) + m_{2} c_{2} x (θ_{2} – θ_{1})

c_{1} = ((m_{3} – m_{2}) x 4200 x (θ_{2} – θ_{1}) + m_{2} c_{2} x (θ_{2} – θ_{1}))/ m_{1 }(100 – θ_{2})

PRECAUTIONS

- Lagging of calorimeter is necessary.
- Thorough stirring of mixtures before taking reading.
- There must be quick transfer of heated metal to the calorimeter.
- Shake off surplus water from hot metal before transferring to the calorimeter.

**EXPERIMENT TO DETERMINE S.H.C OF A LIQUID BY METHOD OF MIXTURE**

Note:

- The S.H.C of solid is known in this case but that of liquid is unknown.
- Liquid different from water is used in the calorimeter.
- Diagram is the same as immediate diagram.

Heat lost by Solid = Heat gained by the liquid + Heat gained by calorimeter and stirrer

m_{1} c_{1 }(100 – θ_{2}) = (m_{3} – m_{2}) x c_{3} x (θ_{2} – θ_{1}) + m_{2} c_{2} x (θ_{2} – θ_{1})

c_{3} = (m_{1} c_{1 }(100 – θ_{2}) – m_{2} c_{2} x (θ_{2} – θ_{1}))/ ((m_{3} – m_{2}) x (θ_{2} – θ_{1}))

Solved Examples

- The hot water tap of a bath delivers water at 85
^{0}c at a rate of 8Kg/min. The cold water tap of the bath delivers water at 30^{0}c at the rate of 20Kg/min. If both taps are left for 5mins, calculate the final temperature of bath water ignoring heat loss to the surrounding. - A piece of copper weighing 400g is heated to 100
^{0}c and then quickly transferred to a copper calorimeter of mass 10g containing 100g of liquid of unknown S.H.C at 30^{0}c. Given that S.H.C of copper is 390 J/Kg/K. If the final temperature of the mixture is 50^{0}c.M

Solution

- Given: m
_{h}= 8Kg/min,M_{h}= 8 x 5 =40Kg; m_{c}= 20 Kg/min, M_{c}= 20 x 5 = 100Kg; θ_{h}= 85^{0}c, θ_{c}= 30^{0}c

Required: θ_{m} =?

Analysis:

Assuming no heat is lost to the surrounding,

Heat lost by hot water = Heat gained by cold water

M_{h} x c_{w} x (θ_{h} – θ_{m}) = M_{c} x c_{w} x (θ_{m} – θ_{c})

40 x (85 – θ_{m}) = 100 x (θ_{m} – 30)

3400 – 40 θ_{m }= 100 θ_{m}– 3000

3400 + 3000 = 100 θ_{m} + 40 θ_{m}

6400 = 140 θ_{m}

Θ_{m }= 6400/140

Θ_{m }= 45.71^{0}c

- Given: m
_{g}= 400g = 0.4Kg; m_{l}= 100g = 0.1Kg; c_{c}=390 J/Kg/K; θ_{1}= 100^{0}C; θ_{2}= 30^{0}C; θ_{3}= 50^{0}C; m_{c}= 10g = 0.01Kg

Required: c_{l} =?

Analysis:

Assuming no heat is lost to the surrounding;

Heat lost by copper block = Heat gained by the liquid + Heat gained by the calorimeter

M_{b} c_{c}(θ_{1}– θ_{3}) = m_{l} x c_{l} x (θ_{3}– θ_{2}) + m_{cal} c_{c} x (θ_{3}– θ_{2})

0.4 x 390 x (100 – 50) = 0.1 x c_{l} x (50 – 30) + 0.01 x 390 x (50 – 30)

7800 = 2c_{l} + 78

7800 – 78 = 2c_{l}

7722 = 2c_{l}

c_{l} = 7722/2 = 3861 J/Kg/K = 3.861 KJ/Kg/K

**ASSIGNMENT**

- A mass M of water at 100
^{0}c is added to another mass, m, of water at a temperature t^{0}c, and the resulting temperature is T^{0}c. If the specific heat capacity of water is 4.2 J/g/K. Derive an expression for T in terms of the other quantities. - The hot water tap of a bath delivers water at 80
^{0}c at a rate of 10g/min. The cold water tap of the bath delivers water at 20^{0}c at a rate of 20kg/min. Assuming that bath taps were left on for 3mins. Find the final temperature of the bath water, ignoring heat losses.

A machine with an input power of 2KW used up 80% of the power. If all the remaining energy appears as heat and heats up 40 Kg of iron, what will be the rise in temperature of this iron in 2mins. (S.H.C of Iron = 500 J/Kg/K).

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