A copper bar of thermal conductivity $401 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\( has one end at \)104^{\circ} \mathrm{C}$ and the other end at \(24^{\circ} \mathrm{C}\). The length of the bar is \(0.10 \mathrm{m}\) and the cross-sectional area is \(1.0 \times 10^{-6} \mathrm{m}^{2} .\) (a) What is the rate of heat conduction, \(\mathscr{P}\) along the bar? (b) What is the temperature gradient in the bar? (c) If two such bars were placed in series (end to end) between the same temperature baths, what would \(\mathscr{P}\) be? (d) If two such bars were placed in parallel (side by side) with the ends in the same temperature baths, what would \(\mathscr{P}\) be? (e) In the series case, what is the temperature at the junction where the bars meet?

Step by step solution

01

Find the rate of heat conduction

Using the given values, we can plug them into the formula for heat conduction: \(\mathscr{P} = kA\frac{(\Delta T)}{d}\). \(k = 401 \mathrm{W}/(\mathrm{m} \cdot \mathrm{K})\) \(A = 1.0 \times 10^{-6} \mathrm{m}^2\) \(\Delta T = 104 - 24 = 80 ^{\circ} \mathrm{C} = 80 \, \mathrm{K}\) \(d = 0.10 \, \mathrm{m}\) Now, solve for \(\mathscr{P}\): \(\mathscr{P} = 401 \cdot 1.0 \times 10^{-6} \cdot \frac{80}{0.10} = \boxed{3.208 \, \mathrm{W}}\)

02

Find the temperature gradient

The temperature gradient is defined as the change in temperature over the length of the bar. To find this, simply divide the change in temperature by the length of the bar: \(\frac{\Delta T}{d} = \frac{80}{0.10} = \boxed{800 \, \mathrm{K/m}}\)

03

Find the rate of heat conduction for series configuration

In the series configuration, two bars are placed end-to-end. This effectively doubles the length of the bar (\(d = 2 \cdot 0.10 = 0.20 \, \mathrm{m}\)), while the cross-sectional area and the thermal conductivity remain the same. Now, we can find the new rate of heat conduction: \(\mathscr{P}_{series} = 401 \cdot 1.0 \times 10^{-6} \cdot \frac{80}{0.20} = \boxed{1.604 \, \mathrm{W}}\)

04

Find the rate of heat conduction for parallel configuration

In the parallel configuration, two bars are placed side by side. This effectively doubles the cross-sectional area (\(A = 2 \cdot 1.0 \times 10^{-6} \, \mathrm{m^2}\)), while the length and the thermal conductivity remain the same. Now, find the new rate of heat conduction: \(\mathscr{P}_{parallel} = 401 \cdot (2 \cdot 1.0 \times 10^{-6}) \cdot \frac{80}{0.10} = \boxed{6.416 \, \mathrm{W}}\)

05

Find the junction temperature in the series case

Let the junction temperature be \(T_{junction}\). We can use the heat conduction formula for the first half of the series configuration: \(1.604 \, \mathrm{W} = 401 \cdot 1.0 \times 10^{-6} \cdot \frac{(104 - T_{junction})}{0.10}\) Now, solve for \(T_{junction}\): \(1.604 \cdot 0.10 = 401 \cdot 1.0 \times 10^{-6} \cdot (104 - T_{junction})\) \(0.1604 = 0.0401 \cdot (104 - T_{junction})\) \(T_{junction} = 104 - \frac{0.1604}{0.0401} = \boxed{64 \, ^{\circ} \mathrm{C}}\)

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