Chapter 17, Problem 13P

To determine

The amount of the energy required to change ice cube from ice to steam.

Answer to Problem 13P

The amount of the energy required to change ice cube from ice to steam is $1.22\times {10}^5\text{J}$.

Explanation of Solution

Given info: The mass of the ice cube is $40.0\text{g}$, the temperature of the ice is $-10.0°\text{C}$ and the temperature of the steam is $110°\text{C}$.

The specific heat of ice is $2090\text{J}/\text{kg}\cdot °\text{C}$, the specific heat of water is $4186\text{J}/\text{kg}\cdot °\text{C}$, the specific heat of steam is $2010\text{J}/\text{kg}\cdot °\text{C}$, the latent heat to melt the ice is $3.33\times {10}^5\text{J}/\text{kg}$, the latent heat to convert water into steam is $2.26\times {10}^6\text{J}/\text{kg}$.

The formula calculate heat required to change the temperature from $-10.0°\text{C}$ to $0°\text{C}$ is,

$Q_1=m{c}_{\text{ice}}\Delta T$

Here,

$m$ is the mass of the ice cube.

$c_{\text{ice}}$ is the specific heat capacity of the ice.

$\Delta T$ the change in the temperature.

Substitute $40.0\text{g}$ for $m$, $2090\text{J}/\text{kg}\cdot °\text{C}$ for $c_{\text{ice}}$ and $\left(0°\text{C}-\left(-10.0°\text{C}\right)\right)$ for $\Delta T$ in above equation to find $Q_1$.

$\begin{array}{c}{Q}_1=\left(40.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(2090\text{J}/\text{kg}\cdot °\text{C}\right)\left(0°\text{C}-\left(-10.0°\text{C}\right)\right)\\ =\left(40.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(2090\text{J}/\text{kg}\cdot °\text{C}\right)\left(10.0°\text{C}\right)\\ =836\text{J}\end{array}$

Thus, the heat required to change the temperature from $-10.0°\text{C}$ to $0°\text{C}$ is $836\text{J}$.

The formula to calculate heat required by the ice to convert into water at constant temperature is,

$Q_{\text{ice}}=m{L}_{\text{f}}$

Here,

$L_{\text{f}}$ is the latent heat required to melt the ice into the water.

Substitute $40.0\text{g}$ for $m$ and $3.33\times {10}^5\text{J}/\text{kg}$ for $L_{\text{f}}$ in above equation to find $Q_{\text{ice}}$.

$\begin{array}{c}{Q}_{\text{ice}}=\left(40.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(3.33\times {10}^5\text{J}/\text{kg}\right)\\ =13320\text{J}\end{array}$

Thus, the heat required by the ice to convert into water is $13320\text{J}$.

The formula calculate heat required to change the temperature from $0°\text{C}$ to $100°\text{C}$ is,

$Q_2=m{c}_{\text{water}}\Delta T$

Here,

$c_{\text{water}}$ is the specific heat capacity of the water.

Substitute $40.0\text{g}$ for $m$, $4186\text{J}/\text{kg}\cdot °\text{C}$ for $c_{\text{water}}$ and $\left(100°\text{C}-0°\text{C}\right)$ for $\Delta T$ in above equation to find $Q_2$.

$\begin{array}{c}{Q}_2=\left(40.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(4186\text{J}/\text{kg}\cdot °\text{C}\right)\left(100°\text{C}-0°\text{C}\right)\\ =16744\text{J}\end{array}$

Thus, the heat required to change the temperature from $0°\text{C}$ to $100°\text{C}$ is $16744\text{J}$.

The formula to calculate heat required by the water to convert into steam at constant temperature is,

$Q_{\text{water}}=m{L}_{\text{v}}$

Here,

$L_{\text{v}}$ is the latent heat required to convert water into steam.

Substitute $40.0\text{g}$ for $m$ and $2.26\times {10}^6\text{J}/\text{kg}$ for $L_{\text{v}}$ in above equation to find $Q_{\text{water}}$.

$\begin{array}{c}{Q}_{\text{water}}=\left(40.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(2.26\times {10}^6\text{J}/\text{kg}\right)\\ =90400\text{J}\end{array}$

Thus, the heat required by the water to convert into steam is $90400\text{J}$.

The formula calculate heat required to change the temperature from $100°\text{C}$ to $110°\text{C}$ is,

$Q_3=m{c}_{\text{steam}}\Delta T$

Here,

$c_{\text{steam}}$ is the specific heat capacity of the steam.

Substitute $40.0\text{g}$ for $m$, $2010\text{J}/\text{kg}\cdot °\text{C}$ for $c_{\text{steam}}$ and $\left(110°\text{C}-100°\text{C}\right)$ for $\Delta T$ in above equation to find $Q_3$.

$\begin{array}{c}{Q}_3=\left(40.0\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(2010\text{J}/\text{kg}\cdot °\text{C}\right)\left(110°\text{C}-100°\text{C}\right)\\ =804\text{J}\end{array}$

Thus, the heat required to change the temperature from $100°\text{C}$ to $110°\text{C}$ is $804\text{J}$.

The total heat needed to convert change ice cube from ice to steam is,

$Q=Q_1+Q_{\text{ice}}+Q_2+Q_{\text{water}}+Q_3$

Substitute $836\text{J}$ for $Q_1$, $13320\text{J}$ for $Q_{\text{ice}}$, $16744\text{J}$ for $Q_2$, $90400\text{J}$ for $Q_{\text{water}}$ and $804\text{J}$ for $Q_3$ in above equation to find $Q$.

$\begin{array}{c}Q=\left(836\text{J}\right)+\left(13320\text{J}\right)+\left(16744\text{J}\right)+\left(90400\text{J}\right)+\left(804\text{J}\right)\\ =1.22\times {10}^5\text{J}\end{array}$

Conclusion:

Therefore, the amount of the energy is required to change ice cube from ice to steam is $1.22\times {10}^5\text{J}$.