Chapter 15, Problem 24P

To determine

To find: The rate of heat out-put.

Answer to Problem 24P

  $1.83\times {10}^3\text{W}$

Explanation of Solution

Given Data:

  $\begin{array}{cc}{T}_H& ={210}^{\circ }\text{C}\\ & ={210}^{\circ }\text{C}+273\\ & =483\text{K}\end{array}$

  $\begin{array}{cc}{T}_L& ={45}^{\circ }\text{C}\\ & ={45}^{\circ }\text{C}+273\\ & =318\text{K}\end{array}$

Output $=950$ W

Formula used:

  $\frac{Q_L}{t}=P_{\text{out}}\left(\frac{T_L}{T_H-T_L}\right)$

Calculation:

  $\begin{array}{l}\frac{Q_L}{t}=P_{\text{out}}\left(\frac{T_L}{T_H-T_L}\right)\\ \begin{array}{l}=(950\text{W})\left(\frac{318\text{K}}{483\text{K}-318\text{K}}\right)\hfill \\ =1.83\times {10}^3\text{W}\hfill \end{array}\end{array}$