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 of temperature is brought to a room of temperature 40. 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 W m⁻¹ ⁻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: (b). Wall area = 2(0.6×0.5 + 0.5×0.2 + 0.6×0.2) = 1.04 m² H = (KAΔ T)/(d) = (0.05×1.04×40)/(0.01) = 208 W Rate of melting = H/L = 2083.4×10⁵ ≈ 6.1×10⁻⁴ kg/s So the answer is (b) 61×10⁻⁵ kg s ⁻¹.

An ice cube of dimensions 60 cm × 50 cm × 20 cm is placed in an… | JEE Main 2022 (26 Jul, Shift 2)

Q31JEE Main 2022 · 26 JUL · SHIFT 2Difficult

Heat Conduction

An ice cube of dimensions $60\text{ cm} \times 50\text{ cm} \times 20\text{ cm}$ is placed in an insulation box of wall thickness $1\text{ cm}$ . The box keeping the ice cube at $0^\circ\text{C}$ of temperature is brought to a room of temperature $40^\circ\text{C}$ . The rate of melting of ice is approximately:

(Latent heat of fusion of ice is $3.4 \times 10^{5}\text{ J kg}^{-1}$ and thermal conducting of insulation wall is $0.05\text{ W m}^{-1}\,{}^\circ\text{C}^{-1}$ )

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Answer: (b)

Wall area $= 2(0.6\times0.5 + 0.5\times0.2 + 0.6\times0.2) = 1.04\ \text{m}^2$
$H = \dfrac{KA\Delta T}{d} = \dfrac{0.05\times1.04\times40}{0.01} = 208$ W
Rate of melting $= \dfrac{H}{L} = \dfrac{208}{3.4\times10^{5}} \approx 6.1\times10^{-4}$ kg/s
So the answer is (b) $61\times10^{-5}$ kg s $^{-1}$ .

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