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 ${70}^{\circ }C$ to ${60}^{\circ }C$ 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

The correct option is C

0.6 Cal/gm ´°C

$S_l=\frac{1}{m_l}\left[\frac{t_l}{t_w}\right(M_w{C}_w+W)-W]$

= $\frac{1}{300}\left[\frac{95}{3\times 60}\right(350\times 1+10)-10]=0.6Cal/gm{\times }^{\circ }C$