Chapter 23, Problem 80A

To determine

The time required to completely vaporize the water in the beaker in seconds.

Answer to Problem 80A

The time required to completely vaporize the water in the beaker is $1.1\times {10}^3\text{ s}$ .

Explanation of Solution

Given:

The mass water is $m=0.10\text{ kg}$ at $25°\text{C}$

The voltage is $V=45\text{V}$ .

The current is $I=5.0\text{ A}$ .

The specific heat of water is $C=4.2\text{kJ}/\text{kg}\cdot °\text{C}$ .

The heat of vaporization of water is $H_{\text{v}}=2.3\times {10}^6\text{J}/\text{kg}$ .

Formula used:

The expression for the heat transfer is,

$Q=mC\Delta T$

Here, $m$ is the mass of an object, $C$ is the specific heat of the object, and $\Delta T$ is the difference between the initial and final temperatures.

The expression for the heat vaporization is,

$Q=m{H}_{\text{v}}$

Here, $m$ is the mass of liquid and $H_{\text{v}}$ is the heat of vaporization of the liquid.

The expression for the power is

$P=IV$

Here, $V$ is the voltage and $I$ is current.

Calculation:

The energy needed to raise the temperature to $100°\text{C}$ is,

$\begin{array}{l}{\left(\Delta Q\right)}_{\text{temperature}}=mC\Delta T\\ {\left(\Delta Q\right)}_{\text{temperature}}=\left(0.10\text{kg}\right)\times \left(4.2\text{kJ}/\text{kg}\cdot °\text{C}\right)\times \left(100°\text{C}-25°\text{C}\right)\\ {\left(\Delta Q\right)}_{\text{temperature}}=32\text{ kJ}\end{array}$

The energy needed to vaporize the water is

$\begin{array}{l}{\left(\Delta Q\right)}_{\text{vaporize}}=m{H}_{\text{v}}\\ {\left(\Delta Q\right)}_{\text{vaporize}}=\left(0.10\text{kg}\right)\times \left(2.3\times {10}^6\text{J}/\text{kg}\right)\\ {\left(\Delta Q\right)}_{\text{vaporize}}=2.3\times {10}^5\text{ J}\times \frac{1\text{ kJ}}{1,000\text{ J}}\\ {\left(\Delta Q\right)}_{\text{vaporize}}=2.3\times {10}^2\text{ kJ}\end{array}$

The total energy needed is

$\begin{array}{l}\Delta Q_{\text{total}}={\left(\Delta Q\right)}_{\text{temperature}}+{\left(\Delta Q\right)}_{\text{vaporize}}\\ \Delta Q_{\text{total}}=\left(32\text{ kJ}\right)+\left(2.3\times {10}^2\text{ kJ}\right)\\ \Delta Q_{\text{total}}=2.6\times {10}^2\text{ kJ}\end{array}$

The power required is,

$\begin{array}{l}P=IV\\ P=\left(\text{5}\text{.0 A}\right)\times \left(\text{45 V}\right)\\ P=225\text{J}/\text{s}\times \frac{1\text{ kJ}}{1,000\text{ J}}\\ P=0.23\text{kJ}/\text{s}\end{array}$

The required to completely vaporize the water in the beaker is,

$\begin{array}{l}t=\frac{\Delta Q_{\text{total}}}{P}\\ t=\frac{2.6\times {10}^2\text{ kJ}}{0.23\text{kJ}/\text{s}}\\ t=1.1\times {10}^3\text{ s}\end{array}$

Conclusion:

Thus, the time required to completely vaporize the water in the beaker is $1.1\times {10}^3\text{ s}$ .