Q149 / 180 · KCET 2023 · Physics
$100 \mathrm{~g}$ of ice at $0^{\circ} \mathrm{C}$ is mixed with $100 \mathrm{~g}$ of water at $100^{\circ} \mathrm{C}$. The final temperature of the mixture is
[Take, $L_f=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ and $\left.S_w=4.2 \times 10^3 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~K}^{-1}\right]$
- A$40^{\circ} \mathrm{C}$
- Correct answerB$10^{\circ} \mathrm{C}$
- C$50^{\circ} \mathrm{C}$
- D$1^{\circ} \mathrm{C}$
Solution
Let the final temperature of mixture be $T$.
Heat absorbed by ice $=$ heat lost by water
$\therefore m L+m c(T-0)=m c(100-T)$
Substituting values,
$\begin{aligned} & \therefore \quad(100 \times 80)+(100 \times 1) T=100 \times 1(100-T) \\ & 8000+100 T=10000-100 T \\ & 200 T=2000 \\ & T=10^{\circ} \mathrm{C} \\ & \end{aligned}$
Hence, the correct answer is option: 'B'.