Two thermal reservoirs are connected by a solid copper bar. The bar is 2 m long,

Question

Two thermal reservoirs are connected by a solid copper bar. The bar is 2 m long,
and the temperatures of the reservoirs are 80.0 °C and 20.0 °C.
a) Suppose the bar has a constant rectangular cross section, 10 cm on a side.
What is the rate of heat flow through the bar?
b) Suppose the bar has a rectangular cross section that gradually widens from the
colder reservoir to the warmer reservoir. The area A is determined by
A = (0.010 m2 )[1.0 + x / (2.0 m)] , where x is the distance along the bar from the
colder reservoir to the warmer one. Find the heat flow and the rate of change of
temperature with distance at the colder end, at the warmer end, and at the
middle of the bar.

Explanation / Answer

we have that P=kA*(Th-Tc)/L. where k of copper = 401. A=0.1*0.1 Th-Tc=60C. L=2m so that P=120(J/s). ------------ we have that. P=kA*dt/dx. where dt is the change of temperature in dx length. so that dT/dx=P/kA we have that dT=P*dx/kA so that dT=(P/401)*dx/(0.01*(1+x/2)) so dT=P*dx/(0.01+0.005x)401 so dT=P*dx*0.5/(2+x) integrate this we have. 60=P*0.5*ln((2+2)/2)=P*0.5*ln2. so that P=173(J/s) ------ and dT/dx=173/(401*0.01*(1+x/2))

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