Topic: Heat Transfer
An air-conditioned room has one of the walls, which is $5m\times3m$ of 3.75 cm thick brick. The conditioned space is maintained at $27^{0}C$ when the outside temperature is $47^{0}C.$ Variation of thermal conductivity with temperature of the wall is given by $k=1+2\times10^{-4}T$ where T is in degree Kelvin, and k is in $W/m^{0}K.$ The heat gained by the conditioned space through this wall is
- A. 849.6W
- B. 84.96W
- C. 800W
- D. 0
Explanation: Given $T_{1}=27^{\circ}C=300~K$, $T_{2}=47^{\circ}C=320~K$. According to Fourier's law, heat transfer $Q=-kA\frac{dT}{dx}$. Area $A = 15~m^{2}$, thickness $dx = 0.0375~m$. Integrating $Q \int dx = -\int A k dT \Rightarrow Q \times 0.0375 = 15 \int_{300}^{320} (1+2\times10^{-4}T) dT$. Evaluating the integral gives $Q = 8496~W$. (Note: Not any suitable answer in given option).