NEET-XII-Physics
28: Heat Transfer
- #5One end of a steel rod (K = 46 J s-1 m-1°C-1) of length 1.0 m is kept in ice at 0°C and the other end is kept in boiling water at 100°C. The area of cross section of the rod is 0.04 cm2. Assuming no heat loss to the atmosphere, find the mass of the ice melting per second. Latent heat of fusion of ice = 3.36 × 105 J kg-1.Ans :
Given:
Thermal conductivity of the rod, K = 46 J s-1 m-1 °C-1
Length of the rod, l = 1 m
Area of the cross-section of the rod, A = 0.04 cm2
= 0.04 × 10-4 m2
= 4 × 10-6 m2
Rate of transfer of heat is given by
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\Delta T}{{\displaystyle \frac{l}{\,\mathrm{\,kA\,}}}}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\frac{\left({T}_{1}-{T}_{2}\right)kA}{l}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=\left(\frac{100-0}{1}\right)\times 46\times 4\times {10}^{-6}``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}=184\times {10}^{-4}\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
`` \,\mathrm{\,Also\,},\,\mathrm{\,\Delta Q\,}=m{\,\mathrm{\,L\,}}_{f}``
`` \therefore \left(\frac{m}{t}\right){L}_{f}=184\times {10}^{-4}``
`` \Rightarrow \left(\frac{m}{t}\right)\times 3.36\times {10}^{5}=184\times {10}^{-4}``
`` \Rightarrow m=\frac{184\times {10}^{-4}}{3.36\times {10}^{5}}\times t``
`` \Rightarrow m=5.5\times 10\times {10}^{-9}\times 1\,\mathrm{\,kg\,}/\,\mathrm{\,s\,}``
`` \Rightarrow m=5.5\times {10}^{-8}\times {10}^{3}\,\mathrm{\,g\,}/\,\mathrm{\,s\,}``
`` \Rightarrow m=5.5\times {10}^{-5}\,\mathrm{\,g\,}/\,\mathrm{\,s\,}``
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