Chapter 15, Problem 23P

To determine

To find: That the temperature of waste heat exhausted.

Answer to Problem 23P

  ${440}^{\circ }\text{C}$

Explanation of Solution

Given Data:

Input $=680$ kcal/sec

Work done $=440$ kw

  $T_H=570$

Formula used:

  $T_{\text{L}}=T_{\text{H}}\left(1-\frac{W}{Q_{\text{H}}}\right)$

Calculation:

  $\begin{array}{l}{T}_{\text{L}}=T_{\text{H}}\left(1-\frac{W}{Q_{\text{H}}}\right)\\ T_{\text{L}}=T_{\text{H}}\left(1-\frac{\left(\frac{W}{t}\right)}{\left(\frac{Q_{\text{H}}}{t}\right)}\right)\\ T_{\text{L}}=(843\text{K})\left(1-\frac{4.40\times {10}^5\text{W}}{2.846\times {10}^6\text{J}/\text{s}}\right)\\ =713K\end{array}$

Converting in centigrade:

  $\begin{array}{cc}{T}_{\text{L}}& =713\text{K}-273\\ & ={440}^{\circ }\text{C}\end{array}$