Q145 / 180 · COMEDK 2024 (Memory Based) · Physics
100 g of ice at $0^{\circ} \mathrm{C}$ is mixed with 100 g of water at $80^{\circ} \mathrm{C}$. The final temperature of the mixture is (Given, latent heat of fusion of ice, $L_f=80 \mathrm{cal} / \mathrm{g}$ and specific heat capacity of water, $S_{\text {water }}=1 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}$ )
- Correct answerA$0^{\circ} \mathrm{C}$
- B$20^{\circ} \mathrm{C}$
- C$40^{\circ} \mathrm{C}$
- D$60^{\circ} \mathrm{C}$
Solution
Let the final temperature of mixture be $T^{\circ} \mathrm{C}$.
From principle of calorimetry,
Heat given by 100 g of water $=$ Heat taken by 100 g of ice
$m_1 S_{\text {water }}\left(\Delta T_1\right)=m_2 L_f+m_2 S_{\text {water }} \Delta T_2$
Substituting the given values, we get
$\begin{aligned}<br/>& 100 \times 1 \times(80-T)=100 \times 80+100 \times 1 \times(T-0) \\<br/>& \Rightarrow \quad 80-T=80+T \\<br/>& \Rightarrow \quad 2 T=0 \Rightarrow T=0^{\circ} \mathrm{C}<br/>\end{aligned}$
From principle of calorimetry,
Heat given by 100 g of water $=$ Heat taken by 100 g of ice
$m_1 S_{\text {water }}\left(\Delta T_1\right)=m_2 L_f+m_2 S_{\text {water }} \Delta T_2$
Substituting the given values, we get
$\begin{aligned}<br/>& 100 \times 1 \times(80-T)=100 \times 80+100 \times 1 \times(T-0) \\<br/>& \Rightarrow \quad 80-T=80+T \\<br/>& \Rightarrow \quad 2 T=0 \Rightarrow T=0^{\circ} \mathrm{C}<br/>\end{aligned}$
Hence, the correct answer is option: 'A'.