# MEASUREMENT OF HEAT ENERGY

Measurement Of Heat Energy:
Heat energy is the energy that is transferred from one place  another as a result of temperature difference. Heat energy always flow from a region of high temperature to the region of lower temperature. Heat energy is called thermal energy. It is measured in joule.

Heat Capacity:
Heat capacity is the quantity of heat energy that is require to raise he temperature of a given mass of an object by one Kelvin. The unit f heat /thermal capacity is joule per Kelvin ( J/K or JK -1 )

Formula Of Heat Capacity, H:
The formula for calculating heat capacity is as stated .
Heat Capacity Cp = mass (m) * Specific heat capacity (c)
Cp = m*c.       ( Cp is also called thermal capacity )

Application of Formula:
Worked Examples:
I. Calculate the heat capacity of a mass 258g if its specific heat capacity is 900JKg-1K-1.

Solution:
Data given in the question:
Mass = 258kg, specific heat capacity, c = 900Jkg-1K-1
Formula:        heat capacity = mass * specific heat capacity
Substitution:               Cp = 258 * 900.  ➡ Cp = 232200JK-1

II. The thermal capacity of an object is 585 JK-1. Calculate the mass of the object if its specific heat capacity is 390 kg-1K-1
Solution:
Data given in the question:
heat capacity =585 JK-1,specific heat capacity is 390 kg-1K-1
Formula: thermal capacity Cp = mass * specific heat capacity
Substitution:              585 = mass * 390
Make mass the subject:  mass = 585 ÷ 390.
➡ mass = 1.5 Kg

III. Determine the specific heat capacity of a substance whose mass is 200 grams and of thermal capacity 250 JK-1

Solution:
Data given in the question:
Mass = 200g = 200÷1000 = 0.2 Kg thermal capacity Cp = 250 JK-1
Formula:  thermal capacity = mass * specific heat capacity
Substitution:            250 = 0.2 * specific heat capacity, c
Make  c the subject:   specific heat  capacity = 250 ÷ 0.2
Specific heat capacity = 1250 JKg-1 K-1.

Specific Heat Capacity:
Specific heat capacity of a substance is the quantity of heat that is required to raise the temperature of 1 kg mass by 1°C or 1Kelvin.
Formula Of Specific Heat Capacity:
The of quantity of  heat that is required to raise the temperature  of a substance by 1 K is directly proportional to the mass of the substance, the temperature change the substance .

Mathematically,
Quantity of heat H ∞ Mass of substance * temperature change
H = c*m*∆@

If we make c the subject t, then we ill get,  c = H/m∆@
H is quantity of heat measured in Joule, m is mass of substance measured in kg, @ is temperature measured in °C or K.
The unit of specific heat capacity is Joule per Kilogram per Kelvin ( JK-1K-1 )

Specific Heat Capacity Of Some Substances:
S/N.    Substance.        Specific heat capacity in JK-1K-1
I. Aluminum.        900
II.  Brass.                 380
III. Copper.             390
IV.   Glass.                 670
VI.  Iron.                   460
VII.  Platinum.          139
VIII. Silver.                234
IX. Tin                     226
X. Zinc.                  384
XI. Alcohol.            2520
XII.  Glycerine.         2430
XIII.  Ice.                    2100
XIV.  Mercury.          140
XV. Paraffin Oil.     2130
XVI. Turpentine.      1760
XVII. Water.              4200

Application Of Formula:
Worked Examples:
I. What is the amount of heat that is required to raise the temperature of 350 g of aluminium cone from 30°C to 68°C if its specific heat capacity is 900 JK-1 K-1.

Solution:
Data given in the question:
Mass = 350 g = 350/1000 = 0.35 kg, s.h.c = 900 JK-1 K-1, @1 = 30°C,
@2 = 68°C. ∆@ = @1 – @2 = 68 – 30 = 38°C
Formula:  Q = mc∆@
Substitution:        Q = 0.35*900*38 ➡ Q = 11970 Joules

II. What is meant by the statement that the specific heat capacity of water is 4200JK-1K-1?
Calculate the temperature change when 1000 J of heat is supplied to 100g of water.

Solution:
Data given in the question:
Mass = 100g = 100/1000 = 0.1 kg, c = 4200JK-1K-1,Q = 1000 J
Formula:      Q = m*c*∆@
Substitution:      1000 = 0.1*4200*∆@
Make @ the subject:   ∆@ = 1000/0.1*4200    ➡ ∆@ = ².38°C

Methods Of Determining The Specific Heat Capacity Of a Substance:
Different methods can be used to determine the specific heat capacity of a substance. The specific heat capacity of a substance can be determined by he following methods:
1. Electrical method.
2. Method of mixtures

Determination Of Specific Heat Capacity Of a Solid By Electrical Method:
In electrical method of determining the specific heat capacity of  a solid substance, electric heater is used to provide the heat required for the experiment.

Aim: to determine the specific heat capacity of a solid.

Apparatus:
solid whose specific heat capacity is o be determined, heater, voltage source, thermometer, calorimeter.
Setup Diagram:

Procedures:
Get a solid whose specific heat capacity you want to determine and that fits into the calorimeter.
Bore two holes in the solid. Weigh and record the mass of the solid. Insert the thermometer and the heater into the holes and add little oil to help establish good thermal contact between the block and the thermometer and heater.
Read and record the initial temperature of the solid and calorimeter. Switch on the electrical heater so that current flows for some times until the temperature rise is about 15°C. Use a stop watch to measure the time for which current flows. Read and record the final temperature of the solid and thermometer.
Data From the Procedures:
Mass of solid = Ms
Mass of calorimeter = Mc
Initial temperature of solid and calorimeter = @1
Final temperature of solid and calorimeter = @2
Voltage applied across heater = V
Current that flow = I
Time for which current flow = t

Theory of calculation:
Heat supplied by heater = heat gained by solid + heat gained by calorimeter.
Formula:
Introducing the formula of the quantity of heat for each of them as stated above, we have:
IVt = Ms*Cs*∆@ + Mc *Cc*∆@
Ms = mass of solid. Cs = specific heat capacity. Mc = mass of calorimeter. Cc  = specific heat capacity of calorimeter. ∆@ = temperature change. I = current.
V = voltage. t = time.
At this point, you make the variable that you want to calculate subject of the formula.

Precautions:
 Make sure that the calorimeter is lagged to prevent heat lose.
 Avoid error due to parallax when taking the reading.

Application Of Formula:-
Worked Examples:
 An electric heater, rated 20V, 40 W, fitted into a metal block supplied heat to the block of mass 2.25 kg and specific heat capacity of 460JK-1K-1. Calculate the temperature rise in the block if the current flow for 10 minutes.

Solution:
Data given in the question:
Voltage = 20 V,  power = IV = 40 W, mass =2.25kg, C = 460 JK-1K-1 ,
time = 10 minutes = 10*60 = 600 seconds.
Note that the mass, specific heat capacity of the calorimeter are not mentioned in the question. Therefore, you have to ignore the quantity of heat aspect of the calorimeter.

Formula:    IVt = Ms*Cs*∆@ + Mc *Cc*∆@
➡ IVt = Ms*Cs*∆@
➡ P*t = Ms*Cs*∆@
Substitution:      40*600 = 2.25*640*∆@
Make ∆@ the subject:      ∆@ = 40*600/2.25*640  ➡ ∆@ = 16.6°C
 An electric heater, rated 20V, 40 W, fitted into a metal block supplied heat to the block of mass M kg and specific heat capacity of 460JK-1K-1. Calculate the value of M if the temperature rise in the block is 25°C and the current flow for 10 minutes.

Solution:
Data given in the question:
Voltage = 20 V,  power = IV = 40 W, mass = Mkg, C = 460 JK-1K-1 , ∆@= 25°C,
time = 10 minutes = 10*60 = 600 seconds.
Note that the mass, specific heat capacity of the calorimeter are not mentioned in the question. Therefore, you have to ignore the quantity of heat aspect of the calorimeter.
Formula:    IVt = Ms*Cs*∆@ + Mc *Cc*∆@  ➡ IVt = Ms*Cs*∆@
➡ P*t = Ms*Cs*∆@
Substitution:      40*600 = Ms*640*25
Make Ms the subject:      Ms = 40*600/25*640  ➡ M = 1.5Kg

Determination Of Specific Heat Capacity Of a Liquid By Electrical Method:
In electrical method of determining the specific heat capacity of  a liquid substance, electric heater is used to heat the liquid during the experiment.

Aim: to determine the specific heat capacity of a liquid.

Apparatus: liquid whose specific heat capacity is o be determined, heater, voltage source, thermometer, calorimeter.
Setup Diagram:

Procedures:
Weigh and record the mass of empty calorimeter. Fill the calorimeter with a liquid whose specific heat capacity you want to determine. Weigh and record the mass of the calorimeter and the liquid. Insert the thermometer and the heater into the liquid in the calorimeter.
Read and record the initial temperature of the liquid and calorimeter. Switch on the electrical heater so that current flows for some times until the temperature rise is about 15°C. Use a stop watch to measure the time for which current flows. Stir the liquid for equal temperature. Read and record the final temperature of the liquid and the calorimeter.

Data From the Procedures:
Mass of empty calorimeter = Mc
Mass of liquid and calorimeter = M LC
Mass of liquid = MLC – MC = ML
Specific heat capacity of liquid = CL
Specific heat capacity of calorimeter = CC
Initial temperature of liquid and calorimeter = @1
Final temperature of liquid and calorimeter = @2
Voltage applied across heater = V
Current that flow = I
Time for which current flow = t

Theory of calculation:
Heat supplied by heater = heat gained by liquid + heat gained by calorimeter.
Formula:
Introducing the formula of the quantity of heat for each of them as stated above, we have:
IVt = ML*CL*∆@ + Mc *Cc*∆@
Make CL the  of formula:     IVt – Mc *Cc*∆@ = ML*CL*∆@
(IVt – Mc *Cc*∆@)/ML*∆@  =  CL
CL  =    IVt – Mc *Cc*∆@/ML*∆@
ML = mass of liquid. Cs = specific heat capacity. Mc = mass of calorimeter. Cc  = specific heat capacity of calorimeter. ∆@ = temperature change. I = current.
V = voltage. t = time.
At this point, you make the variable that you want to calculate subject of the formula.

Precautions:
 Stir the liquid continuously for equal temperature.
 Avoid error due to parallax when taking the reading of the thermometer.
 Make sure the calorimeter is properly lagged.
 Do not use large current or voltage so as not o damage the appliance.

Application Of Formula:
Worked Examples:
350g of water is heated so that its temperature rises from 30°C to 67°C in 35 minutes. Calculate the quantity of heat supplied and the heat supplied per minute. ( s.h.c. of water = 4200 JK-1K -1 )

Solution:
Data given n he question:
Mass = 350g = 350/1000 = 0.35kg, @1= 30°C, @2 = 67°C, Cw = 4200JK-1 K-1 ,
∆@ = @1 – @2 = 67 – 30 = 37°C.
Formula:  Quantity of neat Q = ML*CL*∆@
Substitution:               Q = 0.35*4200*37. ➡ Q = 54390 Joules.
Quantity of heat per minute = total quantity of heat / total time = 54390/7*60 = 129.5 Joules.