A steel bar 10.0 cm long is welded end to end to a copper bar 20.0 cm long. Each

Question

A steel bar 10.0 cm long is welded end to end to a copper bar 20.0 cm long. Each bar has a square cross section 2.00 cm on a side. The free end of the steel bar is in contact with steam at 100 C, and the free end of the copper bar is in contact with ice at 0 C. Find the temperature at the junction of the two bars and the total rate of heat flow.

SOLUTION

SET UP (Figure 1) shows our sketch; we use the subscripts s and c for steel and copper, respectively. The key to the solution is the fact that the heat currents in the two bars must be equal; otherwise, some sections would have more heat flowing in than out, or the reverse, and steady-state conditions could not exist.

SOLVE Let T be the unknown junction temperature; we use the equation H=kA(THTC)/Lfor each bar and equate the two expressions:

ksA(100CT)Ls=kcA(T0C)Lc

The areas A are equal and may be divided out. Substituting numerical values, we find that

[50.2W/(mK)](100CT)0.100m=[385W/(mK)](T0C)0.200m

Rearranging and solving for T, we obtain

T=20.7C

To find the total heat current H, we substitute this value for T back into either of the earlier expressions:

H==[50.2W/(mK)](0.0200m)2(100C20.7C)0.100m15.9W

or

H==[385W/(mK)](0.0200m)2(20.7C0C)0.200m15.9W

REFLECT Even though the steel bar is shorter, the temperature drop across it is much greater than that across the copper bar because steel is a much poorer conductor of heat.

QUESTION: If we change the length of the steel bar, what length is needed in order to make the temperature at the junction 40 C?

Explanation / Answer

Let T be the unknown junction temperature; we use the equation H=kA(THTC)/Lfor each bar and equate the two expressions:

ksA(100CT)/Ls=kcA(T0C)/Lc

The areas A are equal and may be divided out. Substituting numerical values, we find that

[50.2W/(mK)](100C40)/Ls=[385W/(mK)](400C)/0.200m

Ls=0.03912 m or 3.912 cm

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