Question

A mass of $50g$ 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 $100g$ 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 $30g$.)

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

Answer 1

Answer: (D)

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
${\left(\frac{dQ}{dt}\right)}_{water}={\left(\frac{dQ}{dt}\right)}_{liquid}$
$or,\frac{\left(m_w{C}_w+W\right)\Delta T}{t}=\frac{\left(m_{\ell }{C}_{\ell }+W\right)\Delta T}{t}$
(W = water equivalent of the vessel)
$or,m_w{C}_w=m_{\ell }{C}_{\ell }$
$\therefore$ Specific heat of liquid, $C_{\ell }=\frac{m_w{C}_w}{m_{\ell }}$
$=\frac{50\times 1}{100}=0.5kcalkg$