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 ºC^–1 )

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61×10^–5 kg s^–1

30×10^–5 kg s^–1

61×10^–1 kg s^–1

208 kg s^–1

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{dQ}{dt} = \frac{KAΔT}{ℓ} \\ A = 2 (0 .6 \times 0 .5 + 0 .5 \times 0 .2 + 0 .2 \times 0 .6) \end{array}$

$\begin{array}{l} = 2 (0 .3 + 0 .1 + 0 .12) \\ = 2 (0 .4 + 0 .12) \\ = 2 (0 .52) \\ = 1 .04 m^{2} \\ R_{th} = \frac{ℓ}{KA} \Rightarrow \frac{1 \times 10^{- 2}}{0 .05 \times 1 .04} \Rightarrow \frac{10^{- 2}}{0 .052} \\ \frac{dQ}{dt} = \frac{ΔT}{R_{th}} \Rightarrow \frac{40 \times 0 .052}{10^{- 2}} \Rightarrow 2 .08 \times 10^{2} J / s \\ 2 .08 \times 10^{2} = m \times 3 .4 \times 10^{5} \\ m = \frac{2 .08}{3 .4 \times 10^{3}} \Rightarrow 0 .61 \times 10^{- 3} kg / s \\ = 61 \times 10^{- 5} Kg / s \end{array}$