An ice cube of dimensions $60 \mathrm{~cm} \times 50 \mathrm{~cm} \times 20 \mathrm{~cm}$ is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at $0^{\circ} \mathrm{C}$ of temperature is brought to a room of temperature $40^{\circ} \mathrm{C}$. The rate of melting of ice is approximately: (Latent heat of fusion of ice is $3.4 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}$ and thermal conducting of insulation wall is $0.05 \mathrm{Wm}^{-1 \circ} \mathrm{C}^{-1}$)

(B) $61 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}$

(A) $61 \times 10^{-1} \mathrm{~kg} \mathrm{~s}^{-1}$

(C) $208 \mathrm{~kg} \mathrm{~s}^{-1}$

(D) $30 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}$

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Heat & Thermodynamics

An ice cube of dimensions

60 cm \times 50 cm \times 20 cm

is placed in an insulation box of wall thickness 1 cm. The box keeping the ice cube at

0^{\circ} C

of temperature is brought to a room of temperature

4 0^{\circ} C

. The rate of melting of ice is approximately:
(Latent heat of fusion of ice is

3.4 \times 1 0^{5} J kg^{- 1}

and thermal conducting of insulation wall is

0.05 Wm^{- 1 \circ} C^{- 1}

)

Answer: (B)

Total surface area:

A = 2 (0.6 \times 0.5 + 0.5 \times 0.2 + 0.2 \times 0.6) = 2 (0.3 + 0.1 + 0.12) = 1.04

m². Heat flow rate:

\frac{d Q}{d t} = \frac{K A Δ T}{ℓ} = \frac{0.05 \times 1.04 \times 40}{0.01} = 208

J/s. For melting:

\frac{d Q}{d t} = \frac{d m}{d t} \times L

, so

\frac{d m}{d t} = \frac{208}{3.4 \times 1 0 ^{5}} = 0.61 \times 1 0^{- 3} = 61 \times 1 0^{- 5}

kg/s.

JEE Main · 2022 · 26 Jul 2022 · Shift 2 · Single correct option