Grade 11 Physics Note

Calorimetry

Specific heat capacity and heat capacity of solid:

The specific heat of a substance is defined as the amount of heat to be given to the substance of unit mass in order to raise its temperature by 1o. Specific heat capacity of solid can be measured by using the method of mixture. The statement specific heat capacity of water is 4200J kg-1 K-1 means 2400j of heat is required to raise the temperature of 1 kg of water through 1oC.

 

Method of mixture for finding specific heat capacity of a solid:

http://www.schoolphysics.co.uk/age16-19/Thermal%20physics/Heat%20energy/text/Specific_heat_capacity_measurement/images/1.png

Figure17_1; Experimental arrangement for determination of specific heat of solid by the method of mixture

 

The method of mixture is used to determine specific heat of solid according to following procedure.

a. The mass of solid (m) is measured

b. The mass of a calorimeter with stirrer (Mc) is measured

c. Some water is poured in to the calorimeter.

d. The mass of a calorimeter and its content (M1) is measured.

e. Temperature of the water (θ1) in the calorimeter is measured.

f. The solid is placed in a stem jacket by suspending it. After 5-6 min. temperature of solid (θ2) is noted.

g. The mixture of water and the solid is stirred continuously until the temperature of the mixture becomes constant. The temperature of the mixture (θ) is noted.

Let,

 S = specific heat of solid

Sc = specific heat of the material of the calorimeter.

S w = specific heat of water

Now mass of water taken = $({{\rm{m}}_1} - {\rm{\: }}{{\rm{m}}_{\rm{c}}}) = {{\rm{m}}_{\rm{w}}}$

Heat lost by solid in cooling from θ2 to θ

${{\rm{Q}}_1} = {\rm{mS\: }}({\theta _{\begin{array}{*{20}{c}}{2\: }\\\:\end{array}}}$-θ)………1

Heat gained by calorimeter, cooling from θ1 to θ

${{\rm{Q}}_2} = \left( {{{\rm{m}}_{\rm{c}}}{\rm{\: }}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{m}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}{\rm{\: }}} \right)(\theta {\rm{\: \: }}$- θ1) ……2

By the principle of calorimeter,

                                                                 ${{\rm{Q}}_1} - {\rm{\: }}{{\rm{Q}}_2}$

                     ${\rm{s}} = \frac{{\left( {{{\rm{m}}_{\rm{c}}}{\rm{\: }}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{m}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}} \right)\left( {\theta {\rm{\: \: }} - {\rm{\: }}{\theta _1}} \right)}}{{{\rm{m}}\left( {{\theta _{\begin{array}{*{20}{c}}{2\: }\\\:\end{array}}} - {\theta _1}} \right)}}$

 

 

Newton’s law of cooling:

Newton’s law of cooling states that the rate of loss of heat of a liquid is directly proportional to the difference in temperature between the liquid and surroundings. Newton’s law of cooling depends the rate of loss of heat of a liquid is directly proportional to the different in temperature between the liquid and the surroundings.

Figure:

The specific heat of liquid is determined as below,

Let ${{\rm{T}}_{\rm{w}}}{\rm{\: and\: }}{{\rm{t}}_{\rm{l}}}{\rm{\: }}$are the time required for water and liquid respectively to cool from temperature ${{\rm{T}}_2}{\rm{to\: }}{{\rm{T}}_1}$

Let ${{\rm{m}}_{\rm{w}}}{\rm{\: }},{{\rm{m}}_{\rm{l}}}{\rm{\: }},{{\rm{S}}_{\rm{w}}}{\rm{\: }},{\rm{\: }}{{\rm{S}}_{\rm{l}}}{\rm{\: are\: }}$the mass of water, mass of liquid, specific heat of water and specific heat of liquid respectively.

Then, rate of heat lost by calorimeter A and water is,

$\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right) = \frac{{{{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tw}}}} \ldots ..1$

Where, ${{\rm{M}}_{\rm{a}}} = {\rm{mass\: of\: calorimeter\: A}}$

                              ${{\rm{S}}_{\rm{c}}} = {\rm{specific\: heat\: capacity\: of\: calorimeter}}$

Again,

Rate of heat lost by calorimeter B and liquid is,

${\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right)_2} = \frac{{{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tl}}}} \ldots ..2$

Where, ${{\rm{M}}_{\rm{B}}} = {\rm{mass\: of\: calorimeter\: B}}$

According to Newton’s law of cooling,

Hence,

    $\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right) = {\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right)_2}$

     $\frac{{{{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tw}}}} = \frac{{{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tl}}}}$

 

$\frac{{({{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}})\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tw}}}} = \frac{{\left( {{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}} \right)\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tl}}}}$

                                              $\frac{{({{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}){\rm{tl}}}}{{{\rm{tw}}}} = {{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}$

    ${{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}} = \frac{{({{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}})}}{{{\rm{ml}}}}\frac{{{{\rm{t}}_{\rm{l}}}}}{{{{\rm{t}}_{\rm{w}}}}} - \frac{{{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}}}}{{{{\rm{m}}_{\rm{l}}}}}$

Hence we can determine the specific heat capacity of given liquid.

 

 

The difference between specific heat capacity and heat capacity

specific heat capacity

heat capacity

1. The specific heat of a substance is defined as the amount of heat to be given to the substance of unit mass in order to raise its temperature by 1o.

 

2. in SI-system it is measured in ${{\rm{J}}^{ - 1}}{\rm{k}}{{\rm{g}}^{ - 1}}{\rm{K}}$

3. It is denoted by C.

2. The heat capacity is the amount of heat required to raise the temperature of an object or substance one degree.

 

2. in SI-system it is measured in ${{\rm{J}}^{ - 1}}{\rm{k}}$

3. It is denoted by S.

 

 

 

 

The factors on which the heat possessed by a body depends upon

1. Mass

2. Change in temperature

3. The substance and the phase of the substance

 

 

The theory for determining the specific heat of a liquid by the method of cooling:

Newton’s law of cooling states that the rate of loss of heat of a liquid is directly proportional to the difference in temperature between the liquid and surroundings. Newton’s law of cooling depends the rate of loss of heat of a liquid is directly proportional to the different in temperature between the liquid and the surroundings.

Figure:

The specific heat of liquid is determined as below,

Let ${{\rm{T}}_{\rm{w}}}{\rm{\: and\: }}{{\rm{t}}_{\rm{l}}}{\rm{\: }}$are the time required for water and liquid respectively to cool from temperature ${{\rm{T}}_2}{\rm{to\: }}{{\rm{T}}_1}$

Let ${{\rm{m}}_{\rm{w}}}{\rm{\: }},{{\rm{m}}_{\rm{l}}}{\rm{\: }},{{\rm{S}}_{\rm{w}}}{\rm{\: }},{\rm{\: }}{{\rm{S}}_{\rm{l}}}{\rm{\: are\: }}$the mass of water, mass of liquid, specific heat of water and specific heat of liquid respectively.

Then, rate of heat lost by calorimeter A and water is,

$\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right) = \frac{{{{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tw}}}} \ldots ..1$

Where, ${{\rm{M}}_{\rm{a}}} = {\rm{mass\: of\: calorimeter\: A}}$

                              ${{\rm{S}}_{\rm{c}}} = {\rm{specific\: heat\: capacity\: of\: calorimeter}}$

Again,

Rate of heat lost by calorimeter B and liquid is,

${\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right)_2} = \frac{{{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tl}}}} \ldots ..2$

Where, ${{\rm{M}}_{\rm{B}}} = {\rm{mass\: of\: calorimeter\: B}}$

According to Newton’s law of cooling,

Hence,

    $\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right) = {\left( {\frac{{{\rm{dQ}}}}{{{\rm{dt}}}}} \right)_2}$

     $\frac{{{{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tw}}}} = \frac{{{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}}\left( {{{\rm{T}}_2} - {{\rm{T}}_1}} \right) + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tl}}}}$

 

$\frac{{({{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}})\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tw}}}} = \frac{{\left( {{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}} \right)\left( {{{\rm{T}}_2} - {\rm{\: }}{{\rm{T}}_1}} \right)}}{{{\rm{tl}}}}$

                                              $\frac{{({{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}}){\rm{tl}}}}{{{\rm{tw}}}} = {{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}}$

    ${{\rm{M}}_{\rm{l}}}{{\rm{S}}_{\rm{l}}} = \frac{{({{\rm{m}}_{\rm{A}}}{{\rm{S}}_{\rm{c}}} + {\rm{\: }}{{\rm{M}}_{\rm{w}}}{{\rm{S}}_{\rm{w}}})}}{{{\rm{ml}}}}\frac{{{{\rm{t}}_{\rm{l}}}}}{{{{\rm{t}}_{\rm{w}}}}} - \frac{{{{\rm{m}}_{\rm{B}}}{{\rm{S}}_{\rm{c}}}}}{{{{\rm{m}}_{\rm{l}}}}}$

Hence we can determine the specific heat capacity of given liquid.

 

Water is used as heating and cooling agent:

As water has high specific heat capacity. It absorbs more quantity of heat than other because of this Water is used as heating and cooling agent.

 

The specific heat capacity of water is more than that of iron:

The specific heat capacity of water is more than that of iron. The water meets this purpose because of its high specific heat capacity. A large quantity of heat is available by cooling the hot water than that of iron.


A small baby needs more wrapping by woolen clothes that the grownups:

Mass and rate of fall of temperature are inversely related to each other. As the mass of baby is less than the mass of adult man. The baby should loss large amount of heat hence small baby needs more wrapping by woolen clothes that the grownups.

 

 

You can eat hot pudding on a large plate easily but you feel difficulty in a bowl;

You can eat hot pudding on a large plate easily but you feel difficulty in a bowl because rate of cooling is faster for large plate than a bowl because rate of cooling also depends on volume.

 

 

The atmospheric temperature in the cities on sea – cost change very little:

Water has high specific heat capacity. A large quantity of heat is needed for cooling sea. As we know heat energy flows from warmer body to colder body.

 

If you want to cool a coke bottle to cool quickly, you put it in defreeze of a refrigerator:

This is due to the fact that loss of heat by coke bottle depends on the temperature of surrounding as the surrounding temperature in defreeze of refrigerator is smaller than that in outer. 


Go Top