Chapter 12.2, Problem 28PP

To determine

The work done by the expansion valve on water.

Answer to Problem 28PP

The work done by the expansion valve on water is $2.322\times {10}^5\text{ J}$.

Explanation of Solution

Given:

An expansion valve does work on 100 g of water to convert its temperature from $90°\text{C}$ to water vapour at $110°\text{C}$ .

Formula Used:

According to First Law of thermodynamics, the thermal energy change of an object is,

  $\Delta U=Q-W$

Where,

Q is the heat supplied to the object

W is the work done by the object.

Now, when the system is isolated from the surroundings, then thermal energy is,

  $\Delta U=-W$........(1)

Here, the thermal energy consists of energy for temperature changes and energy for converting water into vapor.

The energy for changing water to vapor is given by,

  $Q_v=m{H}_{\text{v}}$

In the above equation, the parameter $H_v$ represents the heat of vaporization of water.

The energy for causing temperature change is given by the expression,

  $Q_t=mC\left(T_B-T_A\right)$

Calculation:

Here, water is having a mass of $m\text{=100 g}$ .The heat of vaporization is $H_{\text{v}}=2.26\times {10}^6\text{ J/kg}$ . The specific heat of water is $C_w=4180\text{ J/kg}\text{.K}$ .The specific heat of vapor is $C_s=2020\text{ J/kg}\text{.K}$

The thermal energy for converting water at $T_1={90}^0\text{C}$ to vapor at $T_2={100}^0\text{C}$ is the sum of thermal energy for converting water to vapor and the energy for causing the temperature change, which is calculated as,

  $\begin{array}{c}{Q}_A=m{C}_w\left(T_2-T_1\right)+m{H}_{\text{v}}\\ =0.100\times 4180\left(100-90\right)+\left(0.100\right)\left(2.26\times {10}^6\right)\\ =2.3018\times {10}^5\text{ J}\end{array}$

Now, thermal energy from vapor at $T_2={100}^0\text{C}$ to vapor at $T_3={110}^0\text{C}$ is given by,

  $\begin{array}{c}{Q}_B=m{C}_s\left(T_3-T_2\right)\\ =0.100\times 2020\left(110-100\right)\\ =2.02\times {10}^3\text{ J}\end{array}$

Therefore, the total energy for the given conversion is given by,

  $\begin{array}{c}\Delta U=Q_A+Q_B\\ =2.3018\times {10}^5\text{ J }+2.02\times {10}^3\text{ J}\\ =2.322\times {10}^5\text{ J}\end{array}$

The work done is therefore equal to this thermal energy and here, work is done by the expansion valve on the object which is water. Thus,

  $\begin{array}{c}\Delta U=-W_{water}\\ =-\left(-W_{valve}\right)\\ =W_{valve}\end{array}$

  $\therefore W_{valve}=2.322\times {10}^5\text{ J}$

Conclusion:

Therefore, the work done by the expansion valve on water is $2.322\times {10}^5\text{ J}$.