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Physics Question on Thermodynamics

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 3.4×1053.4 × 10^5 Jkg–1J kg^{–1 }and thermal conducting of insulation wall is 0.050.05 Wm–1Wm^{–1}°C–1)C^{–1})

A

61×10−3kgs−161 \times 10^{-3}kgs^{-1}

B

61×10−5kgs−161 \times 10^{-5}kgs^{-1}

C

208  kgs−1208\;kgs^{-1}

D

30×10−3kgs−130 \times 10^{-3}kgs^{-1}

Answer

61×10−5kgs−161 \times 10^{-5}kgs^{-1}

Explanation

Solution

**The rate of melting of ice is : **

ΔQΔt=kA(T1−T2)l\frac{Δ_Q}{Δ_t} = \frac{kA(T_1-T_2)}{l}

⇒mLΔt=kA(T1−T2)l⇒ \frac{mL}{Δt} = \frac{kA(T_1-T_2)}{l}

⇒mΔt=kA(T1−T2)Ll⇒ \frac{m}{Δt} = \frac{kA(T_1-T_2)}{Ll}

≈61.1×10−5kg/s\approx 61.1 \times 10^{-5} kg/s

Hence, the correct option is (B): 61×10−5kgs−161 \times 10^{-5}kgs^{-1}