Chapter 18, Problem 40P

To determine

The rate of cooling water pass over the coils if the water temperature is to rise by no more than $7.50C°$ .

Answer to Problem 40P

  $0.13kg/s$

Explanation of Solution

Given info:

Current, $I=17.5Amp$

Voltage, $V=240V$

Change in temperature, $\Delta T=7.50°C$

Formula used:

By the definition of the power,

  $P=VI$

Where, $V$ is voltage and $I$ is current.

Calculation:

By using the formula of power,

  $\begin{array}{l}{P}_{Electric}=P_{Toheatwater}\\ VI=\frac{Q_{Heatwater}}{t}=\frac{mc\Delta T}{t}\end{array}$

  $\Rightarrow \frac{m}{t}=\frac{IV}{c\Delta T}$

Substitute the given values:

  $=\frac{\left(17.5A\right)\left(240V\right)}{\left(4186J/KG.°C\right)\left(7.50°C\right)}=0.134kg/s\approx 0.13kg/s$

Conclusion: The rate is $0.13kg/s$ .