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 -10 C -1)

(A) 61 \times 10^{-1} kg\,s ^{-1}

(D) 30 \times 10^{-5} kg\,s ^{-1}

JEE Mains · Physics · STD 11 - Transmission of heat

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 \(40^{\circ}\,C\). The rate of melting of ice is approximately. (Latent heat of fusion of ice is \(3.4 \times 10^{5}\,J\,kg ^{-1}\) and thermal conducting of insulation wall is \(0.05\,Wm ^{-10} C ^{-1}\) )

Pick an option to check your answer.

Verified Solution

Correct Answer

(B) \(61 \times 10^{-5} kg\,s ^{-1}\)

Detailed explanation

\(\frac{ dQ }{ dt }=\frac{ KA \Delta T }{\ell}\)

\(A =2(0.6 \times 0.5+0.5 \times 0.2+0.2 \times 0.6)\)

\(=2(0.3+0.1+0.12)\)

\(=2(0.4+0.12)\)

\(=2(0.52)\)

\(=1.04\,m ^{2}\)

\(R _{ th }=\frac{\ell}{ KA } \Rightarrow \frac{1 \times 10^{-2}}{0.05 \times 1.04} \Rightarrow \frac{10^{-2}}{0.052}\)

\(\frac{ dQ }{ dt }=\frac{\Delta T }{ R _{ th }} \Rightarrow \frac{40 \times 0.052}{10^{-2}} \Rightarrow 2.08 \times 10^{2}\,J / s\)

\(2.08 \times 10^{2}= m \times 3.4 \times 10^{5}\)

\(m =\frac{2.08}{3.4 \times 10^{3}} \Rightarrow 0.61 \times 10^{-3}\,kg / s\)

\(=61 \times 10^{-5}\,Kg / s\)

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