Chapter 20, Problem 36P

(a)

To determine

To Calculate:The thermal resistance of each cube.

Answer to Problem 36P

Thermal resistance of copper cube is 0.0831 K/W.

Thermal resistance of aluminum cube is 0.141 K/W.

Explanation of Solution

Given: A copper cube and aluminum cube each having3.00cm long edges.

Formula used:

Thermal resistance $R=\frac{\left|\Delta x\right|}{kA}$

  $\left|\Delta x\right|$ is the thickness, k is the conductivity, and A is the area of the cube.

Calculation:

Thermal resistance of copper cube is calculated as follows:

Substitute 3 cm for, $\left|\Delta x\right|$ $\text{401 W/m}\cdot \text{K}$ for K, and ${\left(\text{3 cm}\right)}^{\text{2}}$ for A in the above equation.

  $\begin{array}{l}RCu=\frac{\text{3cm}}{\left(\text{401 W/m}\cdot \text{K}\right){\left(\text{3 cm}\right)}^{\text{2}}}\left(\frac{\text{1cm}}{{\text{1×10}}^{\text{-2}}\text{m}}\right)\\ =0.0831\text{ K/W}\end{array}$

Therefore, thermal resistance of copper cube is 0.0831 K/W.

Thermal resistance of aluminum cube is calculated as follows:

Substitute 3 cm for $\left|\Delta x\right|$ , $\text{237 W/m}\cdot \text{K}$ for K, and ${\left(\text{3 cm}\right)}^{\text{2}}$ for A in the above equation.

  $\begin{array}{l}RAl=\frac{\text{3cm}}{\left(\text{237 W/m}\cdot \text{K}\right){\left(\text{3 cm}\right)}^{\text{2}}}\left(\frac{\text{1cm}}{{\text{1×10}}^{\text{-2}}\text{m}}\right)\\ =0.141\text{ K/W}\end{array}$

Therefore, thermal resistance of aluminum cube is 0.141 K/W

Conclusion: Thermal resistance can be calculated by using heat conductivity.

(b)

To determine

To Calculate:The thermal resistance of two cube combination.

Answer to Problem 36P

The thermal resistance of two-cube combination is 0.224 K/W.

Explanation of Solution

Given:

Cubes are connected in series.

A copper cube and aluminum cube each has 3.00cm long edges.

0.0831 K/W for $R_{Cu}$ and 0.141 K/W for $R_{Al}$

Calculation:

The cubes are in series, so the resultant thermal resistance is:

  ${\mathrm{R}}_{eq}=Rcu+RAI$

Substitute 0.0831 K/W for $R_{Cu}$ and 0.141 K/W for $R_{Al}$ , in the above equation.

  $\begin{array}{l}{\mathrm{R}}_{eq}=0.0831\text{ K/W}+0.141\text{ K/W}\\ =0.224\text{ K/W}\end{array}$

Conclusion: Thus, the thermal resistance of two-cube combination is 0.224 K/W.

(c)

To determine

To Calculate:The thermal current I.

Answer to Problem 36P

Thermal current I at the interface between the two cubes is 0.375 kW.

Explanation of Solution

Given: A copper cube and aluminum cube each has 3.00cm long edges.

  $\Delta T=\left(\text{373 K - 293 K}\right)$

  $R_{eq}=0.224\text{ K/W}$

Calculation:

The expression for the thermal current I is

  $I=\frac{\Delta T}{\mathrm{Re}q}$

Here, ΔT is the difference in temperature and R_eq is the equivalent resistance.

Substitute the values

  $\begin{array}{l}I=\frac{(373\text{K-293K})}{0.224K/W}\\ =357\text{W}\left(\frac{1\text{k}}{1\times {10}^{-3}}\right)\\ =0.357\text{kW}\end{array}$

Conclusion:

Therefore, thermal current I at the interface between the two cubes is 0.375 kW.

(d)

To determine

To Calculate:The temperature at the interface of two cubes.

Answer to Problem 36P

  $\text{70}\text{.3°C}$

Explanation of Solution

Given:

I = 357 W

  $R_{Cu}$ = 0.0831 K/W and $R_{Al}$ = 0.141 K/W

Calculation:

The temperature at the interface between the two cubes is

  $Tinterface=373K-\Delta TCu$

Substitute $I{R}_{Cu}$ for $\Delta TCu$ in the above equation.

  $Tinterface=373K-IRCu$

  $\begin{array}{l}Tinterface=373\text{ K}-\left(\left(\text{357 W}\right)\left(\text{0}\text{.0831 K/W}\right)\right)\\ =343.3\text{ K}\\ =\left(343.3-273\right)°\text{C}\\ =70.3°\text{C}\end{array}$

Conclusion: Therefore, the temperature at the interface between the two cubes is $\text{70}\text{.3°C}\text{.}$