26B - Finding the Specific heat capacity of a liquid using method of cooling.md

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This is using the same methods and the calculations used in the 26A - Verifying newton's cooling law practical, but here we do this for water and another liquid with unknown specific heat capacity.

• Concept

Here since we are using the same calorimeter and same volume of liquid for both the times, the rate of loosing heat ($\frac{dH}{dt}$) will be the same for both.

Using that relation we can find the unknown specific heat capacity.

$$ \begin{align*} \frac{dH}{dt} &= MS\frac{d\theta}{dt}\\\ (M_cS_c + M_wS_w)\frac{d\theta}{dt}_w &= (M_cS_c + M_lS_l)\frac{d\theta}{dt}_l\\ (M_cS_c + M_wS_w)\frac{\delta \theta}{t_w} &= (M_cS_c + M_lS_l)\frac{\delta \theta}{t_l}\\\ \therefore \frac{(M_cS_c + M_wS_w)}{{t_w}} &= \frac{(M_cS_c + M_lS_l)}{t_l}\\ \end{align*} $$

From this we can find $S_l$

Here, $t_w$ is for time it took for water go for $\delta \theta$ temperature difference Here, $t_l$ is for time it took for the liquid go for $\delta \theta$ temperature difference

Important points

All the points in 26A - Verifying newton's cooling law are applied here

• Why do we have to use the same calorimeter for both the experiments?

To make sure the same area of the calorimeter is used to cooling for both - same A
To make sure the cooling constant for both - same K

$$ \begin{align*} \frac{dH}{dt}= KA(\theta_S - \theta_R) \end{align*} $$

• Why do we have to use the same volume of water inside the calorimeter?

To make sure the same conditions are applied for both the times
To make sure the same surface area of the calorimeter is used for cooling conditions