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A principle of calorimetry states that if there is no loss of heat in surrounding the total heat lost by hot body equal to the total heat gained by a cold body.

i.e. heat loss = heat gain

Following are the steps to determine the specific heat capacity of solid from the method of a mixture.

- The solid is first weighted and then is heated such as in a steam chamber of Regnant's apparatus.
- The steam from the boiler is passed through the space between the two walls of the chamber through inlet I and outlet O. A thermometer T
_{1}are inserted to measure the temperature of solid as in the figure. - A calorimeter and stirrer are weighted and half of its volume is filled with water. Then it is again weighted to know the mass of water.
- The calorimeter is placed inside the wooden box by surrounding the calorimeter with woolen clothes to avoid the heat loss.
- The temperature of a calorimeter and its content is measured by the thermometer, T
_{2}. - When a solid gains a constant temperature in the steam chamber, the calorimeter is brought below the chamber, its lid is opened and hot solid is put carefully in the calorimeter.
- The mixture is then stirred gently until the temperature of the mixture reaches a constant value.
- The mass of calorimeter with the stirrer, the mass of water and initial temperature are noted.
- After transferring solid into the calorimeter note the final temperature.

The specific heat of solid is calculated as follows:

m = mass of solid

m_{1} = mass of calorimeter and stirrer

m_{2} = mass of calorimeter, stirrer and water

m_{w} = m_{2}-m_{1} = mass of water

θ_{1}= temperature of solid

θ_{2}= temperature of calorimeter, stirrer and water

θ_{3}= final temperature of mixture

s_{w} = specific heat capacity of water

s_{c} = specific heat capacity of calorimeter and stirrer

s = specific heat capacity of solid

heat gain by solid\( = ms(\theta_1 - \theta )\)

heat gained by water and calorimeter\( = m_ws_w(\theta - \theta_2 ) +m_1s_c(\theta - \theta_2 ) \)

$$= (m_ws_w + m_1s_c)(\theta - \theta_2 )$$

From the principle of calorimetry,

$$ \text {heat gain}= \text {heat loss} $$

$$ms(\theta_1 - \theta ) = (m_ws_w + m_1s_c)(\theta - \theta_2 )$$

$$\therefore s = \frac{(m_ws_w + m_1s_c)(\theta - \theta_2 )}{ms(\theta_1 - \theta )}$$

When a liquid is heated of higher temperature and placed to cool. Then the rate of heat lost by a temperature of the liquid is directly proportional to the difference in temperature of the surrounding.

Let the liquid having mass’ and specific heat capacity ‘s’ is heated to the temperature and placed to cool. Then from Newton’s law of cooling

Rate of heat lost by the liquid \(\frac{dQ}{dt}∝(\theta-\theta_s )\)

$$or,\frac{dQ}{dt}= -K(\theta-\theta_s )$$

Where K is proportionality constant. Its value depends on upon the nature of liquid and its surface area exposed to the surrounding. The negative sign indicates that the quantity of heat lost by liquid goes on decreasing with time.

Consider a liquid of mass ‘m’ and specific heat capacity ‘s’ when ‘dQ’ quantity of heat is supplied to that liquid than the small change in the temperature \(‘d\theta’ \) is increased.

Then we can write,

$$dQ=msd\theta$$

dividing both sides by dt

$$\frac{dQ}{dt} = \frac{msd\theta}{dt}\dots(i)$$

From newton’s law of cooling

$$-K(\theta - \theta_s ) =\frac{dQ}{dt}\dots(ii)$$

From equation (i) and (ii)

$$-K(\theta - \theta_s ) =ms\frac{d\theta}{dt}$$

$$\frac {d\theta}{\theta-\theta_s} =-\frac{K}{ms}dt$$

Integrating on both sides,

$$\int\frac {d\theta}{\theta-\theta_s}=\int\frac{-K}{ms} dt$$

$$log(\theta - \theta_s )=\frac{-K}{ms} t + C\dots(iii)$$

Where ‘C ’ is the constant of integration. The eq.(iii) is an equation of straight line which verifies the Newton's law of cooling.

**Specific Heat of a Liquid by the Method of Cooling**

Newton’s law of cooling can be used to measure the specific heat capacity of a liquid. It is based on the principle that when the two liquids are cooled under identical conditions, their rates of cooling are equal.

Following are the steps to determine specific heat capacity of solid from method of cooling:

- Take two calorimeters A and B of the same materials having masses m
_{1}and m_{2}respectively. - Fill the calorimeter A with some water having mass M
_{1}and calorimeter B with an equal volume of experimental liquid of mass M_{2}. - Now place two calorimeters inside the constant temperature enclosure as shown in the figure.
- The initial temperature is noted.
- The two calorimeters are provided with identical conditions.

If t_{1} and t_{2} be the time taken by water and liquid to cool from to,

Heat lost by the calorimeter \(A = m_1s(\theta_1 - \theta_2 )\)

and Heat lost by the water in calorimeter \(A = m_1s_1(\theta_1 - \theta_2 )\)

where s_{1} is the specific heat capacity of water. Therefore, total heat lost by the calorimeter A and water in cooling from to.

$$ = m_1s(\theta_1 - \theta_2 ) + m_1s_1(\theta_1 - \theta_2 ) $$

$$ = (m_1s + m_1s_1)(\theta_1 - \theta_2 )\dots(i) $$

The rate of cooling of water and calorimeter

$$ = \frac{(m_1s + M_1s_1)}{t_1}(\theta_1 - \theta_2 )\dots(ii) $$

The rate of cooling of liquid and calorimeter B

$$ = \frac{(m_1s + M_2s_1)}{t_2}(\theta_1 - \theta_2 )\dots(iii) $$

Where s_{l} is the specific heat capacity of the liquid

From the law of cooling

Rate of cooling of water and calorimeter A = Rate of cooling of liquid and calorimeter B

$$ \text{or, }\frac{(m_1s + M_1s_1)}{t_1}(\theta_1 - \theta_2 ) =\frac{(m_1s + M_2s_1)}{t_2}(\theta_1 - \theta_2 )$$

$$ \text{or, }\frac{(m_1s + M_1s_1)}{t_1} =\frac{(m_1s + M_2s_1)}{t_2}$$

$$ \text{or, }\frac{(m_1s + M_1s_1)}{m_2} \times \frac{t_2}{t_1} - \frac{M_2}{m_2} s_1\dots(iv) $$

In CGS system, s_{1} = 1calg^{-1}^{-1}. Equation (iv) can be written as

$$ \text{or, }\frac{(m_1s + M_1)}{m_2} \times \frac{t_2}{t_1} - \frac{M_2}{m_2} $$

this equation is used to determine the specific heat capacity of the liquid.

A principle of calorimetry states that if there is no loss of heat in surrounding the total heat loss by hot body equals to total heat gained by a cold body.

Then the rate of heat lost by a temperature of the liquid is directly proportional to the difference in temperature of the surrounding.

Newton’s law of cooling can be used to measure the specific heat capacity of a liquid.

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The rate of cooling of two different liquids under identical condition is same...please make us clear that how this becomes same?

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