Ice of mass 600 g at −10°C is placed in a copper vessel initially at 350°C. At equilibrium 550 g of ice remains mixed with water. Specific heats: copper 0.1 cal g⁻¹ °C⁻¹; ice 0.5 cal g⁻¹ °C⁻¹; latent heat of ice 80 cal g⁻¹. If vessel mass is M×100 g, find M.
Physics · Mixed Chapter
JEE Mains

Question

Ice of mass 600 g at −10°C is placed in a copper vessel initially at 350°C. At equilibrium 550 g of ice remains mixed with water. Specific heats: copper 0.1 cal g⁻¹ °C⁻¹; ice 0.5 cal g⁻¹ °C⁻¹; latent heat of ice 80 cal g⁻¹. If vessel mass is M×100 g, find M.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — Equilibrium temperature: The presence of both ice and water implies final temperature = 0°C. Step 2 — Heat to warm all ice to 0°C: Q₁ = 600×0.5×10 = 3000 cal. Step 3 — Heat to melt 50 g of ice: Melted mass = 600−550=50 g. Q₂ = 50×80=4000 cal. Total gained by ice-water mixture = 7000 cal. Step 4 — Heat lost by the copper vessel: Q(copper) = (100M)×0.1×(350−0)=3500M cal. Step 5 — Set heat loss equal to gain: 3500M=7000 ⇒ M=2. Answer: 2.