NEET-XII-Physics
28: Heat Transfer
- #6A icebox almost completely filled with ice at 0°C is dipped into a large volume of water at 20°C. The box has walls of surface area 2400 cm2, thickness 2.0 mm and thermal conductivity 0.06 W m-1°C-1. Calculate the rate at which the ice melts in the box. Latent heat of fusion of ice = 3.4 × 105 J kg-1.Ans : Area of the walls of the box, A = 2400 cm2 = 2400 × 10-4 m2
Thickness of the ice box, l = 2 mm = 2 × 10-3 m
Thermal conductivity of the material of the box, K = 0.06 W m-1 °C-1
Temperature of the water outside the box, T1 = 20°C
Temperature of ice, T2 = 0°C
`` \,\mathrm{\,Rate\,}\,\mathrm{\,of\,}\,\mathrm{\,flow\,}\,\mathrm{\,of\,}\,\mathrm{\,heat\,}=\frac{\,\mathrm{\,Temperature\,}\,\mathrm{\,difference\,}}{\,\mathrm{\,Thermal\,}\,\mathrm{\,resistance\,}}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{{T}_{1}-{T}_{2}}{{\displaystyle \frac{l}{kA}}}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\left(\frac{20}{2\times {10}^{-3}}\right)\times 0.06\times 2400\times {10}^{-4}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=24\times 6=144\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
Rate at which the ice melts =`` \frac{m{L}_{f}}{t}``
`` \Rightarrow \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\left(\frac{m}{t}\right){L}_{f}``
`` \Rightarrow 144=\left(\frac{m}{t}\right)\times 3.4\times {10}^{5}``
`` \Rightarrow \frac{m}{t}=\frac{144}{3.4\times {10}^{5}}\,\mathrm{\,kg\,}/\,\mathrm{\,s\,}``
`` \Rightarrow \frac{m}{t}=\frac{144\times 60\times 60}{3.4\times {10}^{5}}\,\mathrm{\,kg\,}/\,\mathrm{\,h\,}``
`` \Rightarrow \frac{m}{t}=1.52\,\mathrm{\,kg\,}/\,\mathrm{\,h\,}``
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