Question

Boiling Water at High Pressure. When water is boiled at a pressure of 2.00 atm, the heat of vaporization is $\mathbf{2}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{}\mathbf{J}\mathbf{/}\mathbf{kg}$and the boiling point is 120°C. At this pressure, 1.00 kg of water has a volume of$\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}{\mathbf{m}}^{\mathbf{3}}$, and 1.00 kg of steam has a volume of$\mathbf{0}\mathbf{.}\mathbf{824}\mathbf{m}{\mathbf{}}^{\mathbf{3}}$. (a) Compute the work done when 1.00 kg of steam is formed at this temperature. (b) Compute the increase in internal energy of the water

Short answer

(a) The work done when 1.00 kg of steam is formed at$120°C$ is $1.67\times {10}^{5}J$.

(b) The internal energy of the water is increased by $2.033\times {10}^{6}J$.

Step-by-step solution

Identification of the given information

The given data can be listed below as:

• The heat of vaporization is,$Q=2.20\times {10}^{6}J/kg$.
• The boiled pressure is,$p=2atm\left(\frac{1.013\times {10}^{5}Pa}{1atm}\right)=2.026\times {10}^{5}Pa$.
• The volume of water is,$V_1=1.00\times {10}^{-3}{m}^3$ .
• The volume of steam is,$V_2=0.824m^3$.
• The mass of steam and water is,$m=1kg$.
• The boiling point temperature is, $T=\left(120°C+273\right)K=393K$.

Understanding the change in the internal energy

The internal energy establishes the relation between the heat energy and work done. The difference between two parameters, namely heat energy and work done, provides the change in internal energy.

(a) Calculation of the work done

Now, work done when the water is vaporized by heat at constant pressure is given by:

$W=p\left(V_2-V_1\right)$

Substitute all the values of in the above formula.

role="math" localid="1664279501477" $$\begin{gathered} W=2.026\times {10}^{5}Pa\left(0.824m^3-1.00\times {10}^{-3}{m}^3\right)\left(\frac{1J}{1Pa·m^3}\right) \\ =1.67\times {10}^{5}J \end{gathered}$$

Therefore, the work done is$1.67\times {10}^{5}J$ .

(b) Calculation of the increase in the internal energy of the water

Now, to calculate the increase in internal energy$∆U$ for 1 kg of steam we will apply the first law of thermodynamics.

$∆U=Q-W$

Substitute the values in the values in the above equation.

$$\begin{gathered} ∆U=2.2\times {10}^{6}J-1.67\times {10}^{5}J \\ =2.033\times {10}^{6}J \end{gathered}$$

Therefore, the increase in internal energy is $2.033\times {10}^{6}J$.