Q 11-10-041NEETJEE MainTop questionMedium
$20$ g of ice at $0°$C is mixed with $40$ g of water at $50°$C. The final temperature of the mixture is (latent heat of fusion $= 80$ cal/g)
Answer: (D) $6.67°$C
Concept. First check whether the water can melt all the ice, then share the leftover heat.
Step 1. Heat available. Water cooling to $0°$C: $40 \times 1 \times 50 = 2000$ cal.
Step 2. Heat to melt all the ice. $20 \times 80 = 1600$ cal. All the ice melts, leaving $400$ cal.
Step 3. Final temperature. The $400$ cal warms $60$ g of water:
$$\theta = \frac{400}{60 \times 1} \approx 6.67°\text{C}$$
Answer. $6.67°$C (Option D).
Solution by Sreeraj P, M.Sc Physics