Q133 / 155 · WBJEE 2021 · Physics
$300 \mathrm{gm}$ of water at $25^{\circ} \mathrm{C}$ is added to $100 \mathrm{gm}$ of ice at $0^{\circ} \mathrm{C}$. The final temperature of the mixture is
- A$12.5^{\circ} \mathrm{C}$
- Correct answerB$0^{\circ} \mathrm{C}$
- C$25^{\circ} \mathrm{C}$
- D$50^{\circ} \mathrm{C}$
Solution
Ref. at $0^{\circ} \mathrm{C}$ water
$(\Delta \mathrm{H})$ liberated $=300 \times 1 \times 25=7500 \mathrm{cal}$
$(\Delta \mathrm{H})$ required $=\mathrm{m}_{\text {ice }} \mathrm{L}_{\mathrm{f}}$
$=100 \times 80 \mathrm{cal}=8000 \mathrm{cal}$
as $(\Delta \mathrm{H})_{\text {req. }}>(\Delta \mathrm{H})_{\text {iib }}$ the mixture remains at $0^{\circ} \mathrm{C}$
$\therefore$ Final temperature of mixture will be $0^{\circ} \mathrm{C}$.
Hence, the correct answer is option: 'B'.