Question

In Newton's experiment of cooling, the water equivalent of two similar calorimeters is 10 gm each. They are filled with 350 gm of water and 300 gm of a liquid (equal volumes) separately. The time taken by water and liquid to cool from $70^{o}C$ to $60^{o}C$ is 3 min and 95 sec respectively. The specific heat of the liquid will be

• A. 0.3 Cal/gm ×°C
• B. 0.5 Cal/gm ×°C
• C. 0.6 Cal/gm ×°C
• D. 0.8 Cal/gm ×°C

Answer 1

Answer: (C)

0.6 Cal/gm ×°C

Answer 2

$S_{l} = \frac{1}{m_{l}}\left\lbrack \frac{t_{l}}{t_{W}}(m_{W}C_{W} + W) - W \right\rbrack$

$= \frac{1}{300}\left\lbrack \frac{95}{3 \times 60}(350 \times 1 + 10) - 10 \right\rbrack$=0.6 Cal/gm×°C