Heat Capacity And Specific Heat Capacity SS2 Physics Lesson Note

MEASUREMENT OF HEAT ENERGY

To assess the quantity of heat energy possessed by a body, three quantities are needed. They are:

1. the change in temperature (θ)
2. the specific heat capacity of the body (C)
3. mass of the body (m)

The quantity of heat Q of a body is a product of the three quantities above as expressed by the equation.

Q = mc∆0 …1

It is measured in Joules

Heat Capacity

This is the quantity of heat required to raise the temperature of a substance by one degree.  It is measured in Joules/K.

C = mc               …..…2

SPECIFIC HEAT CAPACITY

The specific heat capacity of a substance is the heat required to raise the temperature of a unit mass of the substance through a one-degree change in temperature.

The quantity of heat Q received by a body is proportional to its mass (m), and temperature change and on the nature of the material the body is made of.

Thus; Q = mc∆0         …..…3

Q = mc(O2 – O1)         …..…4

C is a constant of proportionality called the specific heat capacity of the body, which depends on the nature of the body.

C =   Q

  m(O2 – O1)             ………5

The unit of specific heat capacity is:

J ÷ KgK  

It can be determined by using:

1. the method of mixtures
2. the electrical method

DETERMINATION OF SPECIFIC HEAT CAPACITY BY MIXTURE METHOD 

The solid lead block is weighed on a balance to be Ms.  A lagged calorimeter is dried and weighed to be Mc.  It is then re-weighed to be Mcw when half filled with water.  The initial temperature of the water is taken to be θ1.

The lead block is suspended in boiling water with a temperature θ2 after which it is transferred to the calorimeter and the mixture stirred to maintain a uniform temperature θ3

The specific heat capacity of the lead can be calculated using the fact that heat loss by the lead = heat gained by calorimeter and water.

Given the specific heat capacity of the calorimeter and water to be Cs, Cc and Cw respectively

Heat loss by the lead = heat gained by calorimeter and water

MsCs(θ2 – θ1) = McCc(θ3 – θ1) + (Mcw + Mc)Cw(θ3 – θ1) …..…6

Cc = 

M2Cc(θ3-θ1)+(M3-M2)Cw(θ3-θ1)

                  Ms(θ2-θ3)

……7

DETERMINATION OF SPECIFIC HEAT CAPACITY BY ELECTRICAL METHOD

1. For a solid:

To calculate the specific heat capacity Cb of a solid brass block, we make two holes in a weighted brass block into which a thermometer and a heating element connected to a source of power supply are inserted. Oil is poured into the holes to ensure thermal conductivity.  Assuming no heat is lost to the surroundings, the total amount of electrical heat energy supplied by the coil is equal to the heat gained by the brass.

Heat energy supplied by the coil = Heat gained by the brass

Ivt = MCb∆θ ….…8

Where I = current, V = Voltage, t = time

From Ohm’s law, 

V = IR     ………9

So equation 8,

I²vt = MCb∆θ                   ….…10

V²t = MCb∆θ         …..…11

 R

For a liquid

Heat supplied = Heat gained by liquid + heat gained by calorimeter

Ivt = M1C1b∆θ + McCc∆θ      ……12

CALCULATIONS ON SPECIFIC HEAT CAPACITY

Example

250g of lead at 170°C is dropped into 100g of water at 0°C.  If the final steady temperature is 12°C, calculate the specific heat capacity of lead.  (Cw = 4.2 x 103 Jkg-1k-1)

Solution 

Heat loss by lead = Heat gained by water

= 0.25 × c × (170 – 12) = 0.1 × 4200(12 – 0)

C = 420 × 12

     0.25 × 158

= 127.6J/Kgk

ASSIGNMENT 

1. Define heat capacity.
2. A metal of mass 0.5kg is heated to 1000C, and transferred to a well-lagged calorimeter of heat capacity 80 J/k containing water of heat capacity 420 J/k at 150C.  If the final steady temperature of the mixture is 250C, find the specific heat capacity of the metal
3. Explain the meaning of the statement, that the specific heat capacity of a substance is 777JKg-1K-1