A cubical cell 0.100m on a side is filled with benzene. The top surface is maintained at 25.0°C and the bottom surface is maintained at 15.0°C. After some time, the system will reach a steady state in which the state of the system does not change with time although heat is flowing. Calculate the

Question

A cubical cell 0.100m on a side is filled with benzene. The top surface is maintained at 25.0°C and the bottom surface is maintained at 15.0°C. After some time, the system will reach a steady state in which the state of the system does not change with time although heat is flowing. Calculate the amount of heat flowing through the benzene per hour after a steady state is achieved, neglecting convection.

Accepted Answer

Since the top of the vessel is at a higher temperature than the bottom, we assume that convection can be neglected. The thermal conductivity of pure benzene at 20°C and 1.00 atm pressure is equal to [latex]0.151 J m^{−1} s^{−1} K^{−1}[/latex]. Assuming that the temperature depends only on z, the vertical coordinate, the temperature gradient is

[latex]\frac{dT}{dz} ≈ \frac{ΔT}{Δz} =\frac{10.0 K}{0.100 m} =100 K m^{−1}[/latex]

We have used an average value for the gradient by replacing dT/dz by the quotient of finite differences ΔT/Δz. Fourier’s law now gives the result

[latex]q_{z}=−κ\Bigl(\frac{ΔT}{Δz} \Bigr)=−(0.151 J m^{−1} s^{−1} K^{−1})(100 K m^{−1})=−15.1 J m^{−2} s^{−1}[/latex]

The cross-sectional area is [latex]0.0100m^{2}[/latex], so the total amount of heat flowing in 1.00 hour is

[latex]q=\left|-\Bigl(15.1 J m^{−2} s^{−1}\Bigr) \Bigl(0.0100m^{2}\Bigr) (3600 s) \right| =544J[/latex]

Question 10.2: A cubical cell 0.100m on a side is filled with benzene. The ...

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