Question

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

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

Answer 1

The correct option is D $0.5kcal/kg$

As the surrounding is identical, vessel is identical time taken to cool both water and liquid (from ${30}^{\circ }C$ to ${25}^{\circ }C$) is same 2 minutes.

$\therefore {\left(\frac{dQ}{dt}\right)}_{water}={\left(\frac{dQ}{dt}\right)}_{liquid}$

Or $\frac{(m_w{c}_w+W)\Delta T_1}{t_1}=\frac{(m_l{c}_l+W)\Delta T_2}{t_2}$

$\therefore \Delta T_1=\Delta T_2,t_1=t_2$

(w = water equivalent of the vessel)

or $m_w{c}_w=m_l{c}_l$

$\therefore$ specific heat of liquid,

$C_l=\frac{m_w{c}_w}{m_l}=\frac{50\times 1}{100}=0.5kcal/kg$