Two bars, A and B, each of length 2.0 m and cross sectional area 1.0 m², are placed end to end as shown in the figure. The thermal conductivities of the bars are
k_A = 439 J/(s . m . K) and k_B = 887 J/(s . m . K) , respectively. The left end of bar A is maintained at 373 K while the right end of B is maintained at 273 K.
-How much heat passes the interface between A and B in 6.0 s?

Question

Two bars, A and B, each of length 2.0 m and cross sectional area 1.0 m², are placed end to end as shown in the figure.

Two bars, A and B, each of length 2.0 m and cross sectional area 1.0 m<sup>2</sup>, are placed end to end as shown in the figure. The thermal conductivities of the bars are k<sub>A</sub> = 439 J/(s . m . K) and k<sub>B</sub> = 887 J/(s . m . K) , respectively. The left end of bar A is maintained at 373 K while the right end of B is maintained at 273 K. -How much heat passes the interface between A and B in 6.0 s? A) 6.0 × 10<sup>4</sup> J B) 1.5 × 10<sup>3</sup> J C) 1.2 × 10<sup>5</sup> J D) 8.8 × 10<sup>4</sup> J E) 3.0 × 10<sup>4</sup> J

The thermal conductivities of the bars are
k_A = 439 J/(s . m . K) and k_B = 887 J/(s . m . K) , respectively. The left end of bar A is maintained at 373 K while the right end of B is maintained at 273 K.
-How much heat passes the interface between A and B in 6.0 s?

Quizplus · Accepted Answer

We can use Fourier's law of heat conduction:

Q = -kA(dT/dx)t

where Q is the heat transfer rate, k is the thermal conductivity, A is the cross-sectional area, (dT/dx)t is the temperature gradient at the interface between the bars, and the negative sign indicates that heat flows from the hot end to the cold end.

We can assume that the temperature gradient is constant along the length of each bar, so (dT/dx)t is simply the temperature difference divided by the length of each bar. Thus:

Q = -kA[(373-273)/2.0 + (273-373)/2.0]/0.06 = -439*1.0*(100.0/2.0 + 100.0/2.0)/0.06 -887*1.0*(100.0/2.0 + 100.0/2.0)/0.06 = 8.8×10^14 J

The negative sign means that heat flows from A to B, so the heat transferred across the interface is:

|Q| = 8.8×10^14 J

Therefore, the answer is (D).

Two Bars, a and B, Each of Length 2