NEET-XII-Physics
28: Heat Transfer
- #1A uniform slab of dimension 10 cm × 10 cm × 1 cm is kept between two heat reservoirs at temperatures 10°C and 90°C. The larger surface areas touch the reservoirs. The thermal conductivity of the material is 0.80 W m-1 °C-1. Find the amount of heat flowing through the slab per minute.Ans : Given:
Thermal conductivity of the material, k = 0.80 W m-1 °C-1
Area of the cross section of the slab, A = 100 cm2 = 10-2 m2
Thickness of the slab, Δx = 1 cm = 10-2 m
`` \,\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{\Delta T}{{\displaystyle \frac{\,\mathrm{\,\Delta \,}x}{k·A}}}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{\left(90-10\right)k·\,\mathrm{\,A\,}}{\,\mathrm{\,\Delta \,}x}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=\frac{\left(80\right)\times 0.8\times {10}^{-2}}{{10}^{-2}}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=64\,\mathrm{\,J\,}/\,\mathrm{\,s\,}``
`` \frac{\Delta Q}{\,\mathrm{\,\Delta \,}t}=64\times 60=3840\,\mathrm{\,J\,}/\,\mathrm{\,min\,}``
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