1 g ice at 0°C is converted completely into steam at 100°C. Find heat required (cal), using L𝒇 = 80 cal/g, cwater = 1 cal/(g·°C), Lᵥ = 536 cal/g.
Physics · Mixed Chapter
JEE Main Practice

Question

1 g ice at 0°C is converted completely into steam at 100°C. Find heat required (cal), using L𝒇 = 80 cal/g, cwater = 1 cal/(g·°C), Lᵥ = 536 cal/g.

G1 Physics · Accepted Answer

Step 1: Melt ice: Q₁ = 1×80 = 80 cal. Step 2: Raise water from 0°C to 100°C: Q₂ = 1×1×100 = 100 cal. Step 3: Vaporize at 100°C: Q₃ = 1×536 = 536 cal. Step 4: Total = 80+100+536 = 716 cal.