Chapter 15, Problem 33P

To determine

To find: The volume of water.

Answer to Problem 33P

  $76\text{L}$

Explanation of Solution

Given Data:

  $COP=7$

Power input $=1$ kilowatt

Formula used:

  $V_w=\frac{(\text{COP})W}{L_{\text{fusion}}{\rho }_w}$

Calculation:

The work done to remove the heat from the water is

  $\begin{array}{cc}W& =Pt\\ & =\left(1.0\times {10}^3\text{W}\right)(3600\text{s})\\ & =3.6\times {10}^6\text{J}\end{array}$

Also,

  $\begin{array}{cc}\text{COP}& =\frac{\left(m{L}_{\text{fusion}}\right)}{W}\\ m& =\frac{(\text{COP})W}{L_{\text{fusion}}}\\ V_w{\rho }_w& ={\frac{(\text{COP})W}{L_{\text{fusion}}}}^W\\ V_w& =\frac{(\text{COP})W}{L_{\text{fusion}}{\rho }_w}\end{array}$

Now calculating volume,

  $\begin{array}{cc}{V}_w& =\frac{(7)\left(3.6\times {10}^6\text{J}\right)}{\left(3.33\times {10}^5\text{J}/\text{kg}\right)\left(1.0\times {10}^3\text{kg}/{\text{m}}^3\right)}\\ & =7.6\times {10}^{-2}{\text{m}}^3\\ & =7.6\times {10}^{-2}{\text{m}}^3\left(\frac{1\text{L}}{1\times {10}^{-3}{\text{m}}^3}\right)\\ & =76\text{L}\end{array}$