Question

A refrigerator converts $100 g$ of water at $10^{\circ} C$ into ice at $0^{\circ} C$ in one hour. What will be the quantity of heat removed per minute? Take the specific heat of water is cal $g ^{-1}{ }^{\circ} C ^{-1}$ and latent heat of ice $=80$ cal $g ^{-1}$

• A. 150 cal $\min ^{-1}$
• B. 200 cal $\min ^{-1}$
• C. 250 cal $\min ^{-1}$
• D. 300 cal $\min ^{-1}$

Answer 1

Answer: (A)

Heat drawn in cooling from $10^{\circ} C$ to $0^{\circ} C$
$\Delta Q_{1}=m s \Delta L=1 \times 100 \times 10=1000$ cal
Heat drawn in converting 100 gram water at $0^{\circ} C$ to ice at $0^{\circ} C$
$\Delta Q _{2}= mL =100 \times 80=8000 cal$
Net quantity of heat removed $=8000+1000=9000$
$\frac{\text { Heat removed }}{\text { minute }}=\frac{9000}{60}$
$=150$ cal $\min ^{-1}$