Chapter 15, Problem 68GP

To determine

The amount of water removed.

Answer to Problem 68GP

The amount of water removed is $14.57\text{kg}$ .

Explanation of Solution

Given info:

The room temperature is, $T_r=25°\text{C}=\left(25°\text{C}+273\right)\text{K}=298\text{K}$ .

The condensing temperature is, $T_c=8°\text{C}=\left(8°\text{C}+273\right)\text{K}=281\text{K}$ .

The work done by the humidifier is, $W=600\text{W}=600\text{W}\times \frac{1\text{J}/\text{s}}{1\text{W}}=600\text{J}/\text{s}$ .

The time for dehumidification is, $t=1\text{h}=1\text{h}\times \frac{3600\text{s}}{1\text{h}}=3600\text{s}$ .

Formula Used:

The expression to calculate the amount of water removed is,

  $m=\frac{Wt{T}_c}{L\left(T_r-T_c\right)}$

Here,

L is the heat of condensation and its value is $2.45\times {10}^6\text{J}/\text{kg}$ .

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}m=\frac{\left(600\text{J}/\text{s}\right)\left(3600\text{s}\right)\left(281\text{K}\right)}{\left(2.45\times {10}^6\text{J}/\text{kg}\right)\left(298\text{K}-281\text{K}\right)}\\ =14.57\text{kg}\end{array}$

Conclusion:

Thus, the amount of water removed is $14.57\text{kg}$ .