Q99 / 180 · AIIMS 2015 · Physics
If $10 \mathrm{~g}$ oficeis added to $40 \mathrm{~g}$ of water at $15^{\circ} \mathrm{C}$, then the temperature ofthemixture is (specificheat of water $=4.2 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}$, Latentheatoffusion of ice $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ )
- A$15^{\circ} \mathrm{C}$
- B$12^{\circ} \mathrm{C}$
- C$10^{\circ} \mathrm{C}$
- Correct answerD$0^{\circ} \mathrm{C}$
Solution
Heat lost by water to come from $15^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$ is
$$
\begin{aligned}
H_1=m s \Delta T & =\frac{40}{1000} \times\left(4.2 \times 10^3\right) \times(15-0) \\
& =2520 \mathrm{~J}
\end{aligned}
$$
Heat required to convert $10 \mathrm{~g}$ ice into $10 \mathrm{~g}$ water at $0^{\circ} \mathrm{C}$ is
$$
H_2=m L=\frac{10}{1000} \times\left(3.36 \times 10^5\right)=3360 \mathrm{~J}
$$
Since $H_2>H_1$, so the whole ice will not be converted into water, whereas the temperature of the whole water will be $0^{\circ} \mathrm{C}$.
Therefore the temperature of the mixture is $0^{\circ} \mathrm{C}$.
Hence, the correct answer is option: 'D'.