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
V. Lead.                  129
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.
 Take the reading when the mercury thread is steady.
 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.
 Take the reading of the thermometer when the mercury thread  is steady.
 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.


Exercises:
 What is meant by the caloric value of a solid fuel. Explain thermal capacity and water equivalent
 Differentiate between specific heat and calorie, heat capacity and specific heat capacity of a substance.
 32g of water at 60°C is poured into 60g of cold  water at 15°C, which is contained in a calorimeter of mass 40g, and s.h.c of 0.1 Jg-1 K-1 . Neglecting heat loss show that the resulting temperature is 29°C
 A piece of metal f mass mass50g heated to 100°C is dropped into 200g of water at 14°C in copper calorimeter weighing 80g and rises the temperature  to 70°C. What is the heat capacity of the piece of metal? ( s.h.c of =0.4J/gk. )
 Some hot water three times added to us mass of water at 10°C and the temperature was 20°C. What was the tempera of the hot water?
 Apiece  mass 300g at a temperature of 950°C  quickly transfered to  vessel of negligible  thermal  capacity containing 250gof water at 25°C . if the final temperature of the mixture is 100°C ,  calculate the mass of water that will boil away. ( Cw = 4200J/kgk, s.h.c. f vaporization of steam = 2.26*106J/kg )
 What is meant by the statement the specific heat capacity of a substance is 400J/kgk?
 Describe an experiment to determine the specific heat capacity of copper, using a copper ball. State ii precautions necessary to obtain accurate results in the experiment.
 A copper ball of mass, 34g at 130°C is placed in a copper calorimeter of mass 60g containing 50g of water at 30°C. Calculate the final temperature of the mixture, ignore heat loss. ( Cw = 4200J/KgK,
Cc = 0.4J/KgK )
 A piece of copper ball of mass 25g at 230°C is placed in a copper calorimeter of mass 60g containing water of mass 54g at 31°C. Calculate that final temperature of the mixture. ( Cc = 400J/KgK, Cw  = 4200J/KgK )
 A tap supplies water at 26°C while another supplies at 83°C. If a man wishes to bath with water at 40°C, what is the ratio to mass of hot water to that of cold water? ( ssce, 19191 ).
 A waterfall is 240m high, calculate the difference in temperature of water between the top and the bottom of the waterfall. Neglect heat loss. ( g = 10m/s², Cw = 4200J/KgK.
 Hot water at a temperature t is added to twice the amount of water at a temperature of 3°C, calculate t.
 A body of specific heat capacity 450J/KgK falls to the ground from rest through a vertical height of 20m. Assuming conservation of energy, calculate the change in temperature of the body in hitting the ground. ( g = 10 m/s²)
 An iron rod of mass 2kg at a temperature of 280°C is dropped into some quantity of water initially at a temperature of 30°C. If the temperature of the mixture is 70°C, calculate the mass of the water. Neglect heat loss. ( Cw = 4200J/KgK, , Ci  = 4200J/KgK )
 When two objects p and q are supplied with same quantity of heat, the temperature change in p is twice that of q. If the masses of p and q are the same, calculate the ratio of  the specific heat capacities of q to p.
 A 400 W immersion heater is used to heat water of mass 0.5kg. If the temperature of the liquid increased by  2.5°C in one second, calculate the specific heat capacity of the liquid.
 How much heat is given out when a piece of iron of mass 50g and specific heat capacity 460J/KgK cool from 85°C 25°C?.
 How long will it take to heat 3kg of water from 28°C to 78°C in an  electric kettle taking a current of 6A from an e.m.f sourec of 240V?
 200g of water at 100°C is poured into 50g of water at 30°C. Assuming heat absorbed by container is negligible, calculate the final temperature of the mixture. Neglecting heat loss.
 A metal of mass 1.5kg was heated form 27°C  47°C in 4 minutes by a boiler ring of 75W rating. Calculate the specific heat  capacity of the metal. Neglect heat loss.
 How long does it take a 750W heater operating at full rating to raise the temperature of 1kg of water from 40°C to 70°C? ( Cw = 4300J/gK )
 A tap supplies water at 30°c while another supplies at 86°C. If a man wishes to bath with water at 44°C, calculate the ratios of the mass of hot water to that of cold water required.
 500g of water is heated so that its temperature rises from 30°C to 72°C in 7 minutes. Calculate the heat supplied per minute. ( Cw = 4300J/KgK )
 A piece of copper of mass 30g losses 60J of heat energy. If the specific heat capacity of copper is 400J/KgK, calculate the change in temperature of the copper.
 A body of mass 20g and specific heat capacity, 400J/KgK absorbs 20 J of heat energy. Calculate its rise in temperature.
 Water of mass 4kg at 30°C absorbs  8.4x105 J of heat. Neglecting heat loss, calculate the final temperature of the mixture. (Cw  = 4200J/KgK )
 Two samples, of pure naphthalene of masses, 100 g and 200 g are heated through temperature difference of 10°C and 20°C respectively. What is the ratio of the specific heat capacity of the small sample to that of the bigger sample?
 The temperature of a piece of metal of mass 9g is raised from 10°C to 110°C when it absorbs 108J of heat energy. Determine the specific heat capacity of the metal.
 Calculate the power rating of an immersion heater used for 10minutes to increase the temperature of 10kg of water by 15K. ( Cw = 4200J/KgK )
 A piece of metal 50g is cooled from 80°C to 20°C. Calculate the amount of heat loss. ( Cm = 450J/KgK )
 400g of water is added to 200g of water at 70°C. If they are properly mixed and the temperature of the mixture is 30°C, calculate the initial temperature of the cold water. Neglect heat absorbed by container.