JEE Main 2022 · 26 July · evening · Single correct · +4 / −1
An ice cube of dimensions 60 cm × 50 cm × 20 cm is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at 0° C of temperature is brought to a room of temperature 40° C. The rate of melting of ice is approximately :
(Latent heat of fusion of ice is 3.4 × 10⁵ J kg⁻¹ and thermal conducting of insulation wall is 0.05 Wm^(-1 °) C⁻¹ )
- (a) 61 × 10⁻³ kg s⁻¹
- (b) 61 × 10⁻⁵ kg s⁻¹
- (c) 208 kg s⁻¹
- (d) 30 × 10⁻⁵ kg s⁻¹
Answer and solution
Answer (b) 61 × 10⁻⁵ kg s⁻¹
Solution
Physics sir · 3 steps
Heat leaks in through all six walls of the box, and every joule that gets in melts some ice.
- Area of the walls. Heat enters through all six faces. In metres the box is 0.6 × 0.5 × 0.2, so A = 2(0.6 × 0.5 + 0.5 × 0.2 + 0.6 × 0.2) = 2 × 0.52 = 1.04 m².
dQ/dt = (kA Δ T)/d = (0.05 × 1.04 × 40)/0.01 = 208 W
- Heat to melting. Each second, 208 J enters and melts a mass m of ice, with mL = 208. So dm/dt = 208/(3.4 × 10⁵) ≈ 6.1 × 10⁻⁴ kg s⁻¹.
- So. The rate of melting is about 61 × 10⁻⁵ kg s⁻¹, which is option (b).
The idea to keep. For a melting problem, find the heat flow in watts with kAΔ T/d, then divide by the latent heat. Count every face that heat can pass through.
Answer from the NTA final answer key. Solution by PrepWiser.