Chapter 15, Problem 54GP

(a)

To determine

The work done by the refrigerator to change the water into ice.

Answer to Problem 54GP

The work done by the refrigerator to change the water into ice is $3.74\times {10}^4\text{J}$ .

Explanation of Solution

Given info:

The temperature of the freezer is, $T_1=-17°\text{C}=\left(-17°\text{C}+273\right)\text{K}=256\text{K}$ .

The temperature of the room is, $T_2=25°\text{C}=\left(25°\text{C}+273\right)\text{K}=298\text{K}$ .

The mass of the water is, $m=0.50\text{kg}$ .

Formula Used:

The expression to calculate the heat removed from the water to change it into ice is,

  $Q_r=m\left(c_l\left(T_i-T_1\right)+L+c_i\left(T_2-T_i\right)\right)$

Here,

• $c_l$ is the specific heat of the water and its value is $4186\text{J}/\text{kg}\cdot \text{K}$ .
• $c_i$ isthe specific heat of the ice and its value is $2100\text{J}/\text{kg}\cdot \text{K}$ .
• L is the latent heat of the ice and its value is $3.33\times {10}^5\text{J}/\text{kg}$ .
• $T_i$is the freezing point of the ice and its value is 273 K.

The expression to calculate thework done by the refrigerator to change the water into ice is,

  $W_1=Q_r\left(\frac{T_2}{T_1}-1\right)$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}{Q}_r=\left(0.50\text{kg}\right)\left[\left(4186\text{J}/\text{kg}\cdot \text{K}\right)\left(273\text{K}-256\text{K}\right)+3.33\times {10}^5\text{J}/\text{kg}+\left(2100\text{J}/\text{kg}\cdot \text{K}\right)\left(298\text{K}-273\text{K}\right)\right]\\ =2.28\times {10}^5\text{J}\end{array}$ Substitute all the values in the above expression.

  $\begin{array}{c}{W}_1=\left(2.28\times {10}^5\text{J}\right)\left(\frac{298\text{K}}{256\text{K}}-1\right)\\ =3.74\times {10}^4\text{J}\end{array}$

Conclusion:

Thus, the work done by the refrigerator to change the water into ice is $3.74\times {10}^4\text{J}$ .

(b)

To determine

The minimum time to change the water into ice.

Answer to Problem 54GP

The minimum time to change the water into ice is $178.09\text{s}$ .

Explanation of Solution

Given info:

The output of the compressor is, $P=210\text{W}=210\text{J}/\text{s}$ .

Formula Used:

The expression to calculate the minimum time to change the water into ice is,

  $t=\frac{W_r}{P}$

Calculation:

Substitute all the values in the above expression.

  $\begin{array}{c}t=\frac{3.74\times {10}^4\text{J}}{210\text{J}/\text{s}}\\ =178.09\text{s}\end{array}$

Conclusion:

Thus, the minimum time to change the water into ice is $178.09\text{s}$ .