NEET-XII-Physics
28: Heat Transfer
- #13Figure (28-E1) shows water in a container having 2.0 mm thick walls made of a material of thermal conductivity 0.50 W m-1°C-1. The container is kept in a melting-ice bath at 0°C. The total surface area in contact with water is 0.05 m2. A wheel is clamped inside the water and is coupled to a block of mass M as shown in the figure. As the block goes down, the wheel rotates. It is found that after some time a steady state is reached in which the block goes down with a constant speed of 10 cm s-1 and the temperature of the water remains constant at 1.0°C. Find the mass M of the block. Assume that the heat flows out of the water only through the walls in contact. Take g = 10 m s-2.
Figure
Ans :
Temperature of water, T1 = 1°C
Temperature if ice bath, T2 = 0°C
Thermal conductivity, K = 0.5 W/m °C
Length through which heat is lost, l = 2 mm = 2 × 10-3 m
Area of cross section, A = 5 × 10-2 m2
Velocity of the block, v = 10 cm/sec = 0.1 m/s
Let the mass of the block be m.
Power = F · v
= (mg) v ......(i)
Also,
`` \,\mathrm{\,Power\,}=\frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}t}.......\left(\,\mathrm{\,ii\,}\right)``
`` \frac{\,\mathrm{\,\Delta Q\,}}{\,\mathrm{\,\Delta \,}\,\mathrm{\,t\,}}=\frac{\,\mathrm{\,k\,}·\,\mathrm{\,A\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{\,\mathrm{\,l\,}}........\left(\,\mathrm{\,iii\,}\right)``
`` ``
From equation (i), (ii) and (iii), we get
`` \left(\mathit{m}\mathit{g}\right)v=\frac{\,\mathrm{\,k\,}·\,\mathrm{\,A\,}\left({\,\mathrm{\,T\,}}_{1}-{\,\mathrm{\,T\,}}_{2}\right)}{l}``
`` m=\frac{0.5\times 5\times {10}^{-2}\left(1\right)}{\left(2\times {10}^{-3}\right)\times 10\times 0.1}``
`` m=12.5\,\mathrm{\,kg\,}``
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