Question

A mass of $50\,g$ of water in a closed vessel, with surroundings at a constant temperature takes $2$ minutes to cool from $30^\circ C$ to $25^\circ C$. A mass of $100\,g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from $30^\circ C$ to $25^\circ C$. The specific heat of the liquid is: (The water equivalent of the vessel is $30\,g$.)

• A. 2.0 kcal/kg
• B. 7 kcal/kg
• C. 3 kcal/kg
• D. 0.5 kcal/kg

Answer 1

Answer: (D)

0.5 kcal/kg

Answer 2

As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from $30?C$ to $25?C$ is same $2$ minutes, therefore
$\left(\frac{d Q}{d t}\right)_{water} =\left(\frac{d Q}{d t}\right)_{liquid}$
$or, \, \frac{\left(m_{w} C_{w}+W\right)\Delta T}{t}=\frac{\left(m_{\ell}C_{\ell}+W\right)\Delta T}{t}$
(W = water equivalent of the vessel)
$or, \, m_{w} C_{w} =m_{\ell} C_{\ell}$
$\therefore$ Specific heat of liquid, $C_{\ell} =\frac{m _{w} C_{w}}{m_{\ell}}$
$=\frac{50\times1}{100}=0.5 \, kcal \, kg$