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  • An ice cube of dimensions <img alt="\mathrm{60 \mathrm{~cm} \times 50 \mathrm{~cm} \times 20 \mathrm{~cm}}" src="/latex-image/?%5Cmathrm%7B60%20%5Cmathrm%7B%7Ecm%7D%20%5Ctimes%2050%20%5Cmathrm%7B%7

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

 (Latent heat of fusion of ice is \mathrm{3.4 \times 10^{5} \mathrm{~J} \mathrm{~kg}^{-1}} and thermal conducting of insulation wall is \mathrm{0.05\; \mathrm{Wm}^{-1 \circ} \mathrm{C}^{-1}}) 

Option: 1

\mathrm{61 \times 10^{-3} \mathrm{~kg} \mathrm{~s}^{-1}}


Option: 2

\mathrm{61 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}}


Option: 3

\mathrm{208 \mathrm{~kg} \mathrm{~s}^{-1}}


Option: 4

\mathrm{30 \times 10^{-5} \mathrm{~kg} \mathrm{~s}^{-1}}


Answers (1)

best_answer

Volume of ice cube \mathrm{=6\times 10^{4}\left ( cm \right )^{3}}

                              \mathrm{=6\times 10^{-2}m^{3}}

\mathrm{Rate \: of \: fusion =\frac{d Q}{d t}=\left(\frac{d m}{d t}\right) L_f}

\mathrm{Rate \: of\: heat\: fiow =\frac{K A(\Delta T)}{t}}

\mathrm{t \rightarrow thickness\: of\: wall}

\mathrm{At\: equilibrium,}

\mathrm{Rate \: of \: fusion = Rate \: of \: heat \: flow}

\mathrm{\left. (\frac{dm}{d t}\right) L_f=\frac{2 K A_1 \Delta T}{\Delta t}+\frac{2 K A_2 \Delta T}{\Delta t}+\frac{2KA_{3}\Delta T}{\Delta t}}

\mathrm{\left(\frac{d m}{d t}\right) \times 3.4 \times 10^S=\left(\frac{2 \times 0.05 \times 40}{10^{-2}}\right)\left[A_1+A_2+A_3\right]}

\mathrm{\left(\frac{d m}{d t}\right)=\frac{4 \times 10^2}{3.4 \times 10^5} \times[3000+1000+1200] \times 10^{-4}}

\mathrm{=\frac{4}{3.4} \times 5.2 \times 10^{-4}=61 \times 10^{-5} \frac{\mathrm{kg}}{\mathrm{s}}}

Hence (2) is correct option.

Posted by

Rishi

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