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
Heat available from the water cooling to $0°$C: $40 \times 50 = 2000$ cal. Heat to melt all the ice: $20 \times 80 = 1600$ cal. All the ice melts, leaving $400$ cal to warm $60$ g of water:
$$\theta = \frac{400}{60} \approx 6.67°\text{C}$$
Solution by Sreeraj P, M.Sc Physics