Chapter 5, Problem 83P

To determine

The temperature of the device at the end of the 5-min operating period with and without the heat sink.

Explanation of Solution

Given:

The rate of electrical work to be done into the system $\left({\dot{W}}_{\text{e,in}}\right)$ is $25\text{W}$.

The mass of the electronic device $\left(m_{\text{device}}\right)$ is 20 g.

The specific heat of constant pressure for electronic device $\left(c_p\right)$ is $850\text{}\text{J}/\text{kg}°\text{C}$.

The change in time $\left(\Delta t\right)$ is 5 min.

The initial temperature of the system $\left(T_1\right)$ is $25°\text{C}$.

Calculation:

Write the expression for the energy balance equation.

  $E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$        (I)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Simplify Equation (I) and write energy balance electrical device without heat sink.

  $\begin{array}{c}{W}_{\text{e,in}}=\Delta U_{\text{device}}\\ =m\left(u_2-u_1\right)\end{array}$        (II)

Here, the electrical work to be done into the system is $W_{\text{e,in}}$, the change in the internal energy of device is $\Delta U_{\text{device}}$, the mass of the electronic device is $m$, the initial specific internal energy of device is $u_1$, and the final specific internal energy of device is $u_2$.

Rewrite the Equation (II) into per unit time.

  ${\dot{W}}_{\text{e,in}}\Delta t=m{c}_p\left(T_2-T_1\right)$

  $\begin{array}{l}\left(25\text{W}\right)\left(5\min \right)=\left(20\text{g}\right)\left(850\text{}\text{J}/\text{kg}°\text{C}\right)\left(T_2-25°\text{C}\right)\\ \left(25\text{W}\times \left(\frac{1\text{J}/\text{s}}{1\text{W}}\right)\right)\left(5\min \times \left(\frac{60\sec }{1\min }\right)\right)=\left[\begin{array}{l}\left(20\text{g}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\\ \left(850\text{}\text{J}/\text{kg}°\text{C}\right)\left(T_2-25°\text{C}\right)\end{array}\right]\\ \left(7500\text{J}\right)=\left(17\text{}\text{J}/°\text{C}\right)\left(T_2-25°\text{C}\right)\\ T_2=466\text{°C}\end{array}$

Thus, the temperature of the device at the end of the 5-min operating period without the heat sink is $\underset{\_}{466°\text{C}}$.

Simplify Equation (I) and write energy balance electrical device with heat sink.

  $\begin{array}{c}{W}_{\text{e,in}}=\Delta U_{\text{device}}+\Delta U_{\text{heat}\text{sink}}\\ =m{\left(u_2-u_1\right)}_{\text{device}}+m{\left(u_2-u_1\right)}_{\text{heat}\text{sink}}\end{array}$        (III)

Here, the electrical work to be done into the system is $W_{\text{e,in}}$, the change in the internal energy of device is $\Delta U_{\text{device}}$, the change in the internal energy of heat sink is $\Delta U_{\text{heat sink}}$, the mass of the electronic device is $m$, the initial specific internal energy of device is $u_1$, the final specific internal energy of device is $u_2$, the initial specific internal energy of heat sink is $u_1$, the final specific internal energy of heat sink is $u_2$.

Rewrite the Equation (II) into per unit time.

  ${\dot{W}}_{\text{e,in}}\Delta t=m{c}_p{\left(T_2-T_1\right)}_{\text{device}}+m{c}_p{\left(T_2-T_1\right)}_{\text{heat sink}}$

From the Table A-3(b), “Properties of common liquids, solids, and foods”, obtain the value of specific heat of constant pressure for aluminium at 300 K temperature is $0.902\text{k}\text{}\text{J}/\text{kg}\cdot °\text{C}$.

Substitute ${\dot{W}}_{\text{e,in}}=25\text{W}$, $\Delta t= 5min$, $m_{\text{device}}=20g$, $c_{p_{d\text{evice}}}=850\text{}\text{J}/\text{kg}°\text{C}$, $T_1=25 C$, $c_{p_{d\text{evice}}}=0.902\text{k}\text{}\text{J}/\text{kg}\cdot °\text{C}$, and $m_{\text{heat}\text{sink}}=0.5kg$ in the above Equation

  $\begin{array}{l}\left(25\text{W}\right)\left(5\min \right)=\left(20\text{g}\right)\left(850\text{}\text{J}/\text{kg}°\text{C}\right)\left(T_2-25°\text{C}\right)+\left(0.5\text{kg}\right)\left(0.902\text{k}\text{}\text{J}/\text{kg}\cdot °\text{C}\right)\left(T_2-25°\text{C}\right)\\ \left(25\text{W}\times \left(\frac{1\text{J}/\text{s}}{1\text{W}}\right)\right)\left(5\min \times \left(\frac{60\sec }{1\min }\right)\right)=\left[\begin{array}{l}\left[\begin{array}{l}\left(20\text{g}\times \left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\\ \left(850\text{}\text{J}/\text{kg}°\text{C}\right)\left(T_2-25°\text{C}\right)\end{array}\right]\\ +\left[\begin{array}{l}\left(0.5\text{kg}\right)\\ \left(0.902\text{k}\text{}\text{J}/\text{kg}\cdot °\text{C}\times \left(\frac{1000\text{}\text{J}/\text{kg}°\text{C}}{1\text{k}\text{}\text{J}/\text{kg}\cdot °\text{C}}\right)\right)\\ \left(T_2-25°\text{C}\right)\end{array}\right]\end{array}\right]\\ \left(7500\text{J}\right)=\left(17\text{}\text{J}/°\text{C}\right)\left(T_2-25°\text{C}\right)+\left(451\text{}\text{J}/°\text{C}\right)\left(T_2-25°\text{C}\right)\\ T_2=41.0\text{°C}\end{array}$

Thus, the temperature of the device at the end of the 5-min operating period with the heat sink is $\underset{\_}{41.0°\text{C}}$.