Chapter 5, Problem 87E

Water at 50°C can be obtained by mixing together which one or more of the following? Which of the others would have final temperatures higher than 50°C and which lower than 50°C?

1. a. 1 kg of ice at 0°C and 1 kg of steam at 100°C
2. b. 1 kg of ice at 0°C and 1 kg of water at 100°C
3. c. 1 kg of water at 0°C and 1 kg of steam at 100°C
4. d. 1 kg of water at 0°C and 1 kg of water at 100°C

To determine

Whether mixing of given samples will result water having temperature equal to $50\text{°C}$ or a mixture having final temperature less than or greater than $50\text{°C}$.

Answer to Problem 87E

The mixing $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$ will result final temperature higher than $50\text{°C}$.

Explanation of Solution

Given info: The samples are $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$.

Write the expression for the heat energy absorbed when ice at $0\text{°C}$ melts into water at $0\text{°C}$.

$Q_1=m{L}_{\text{f}}$

Here,

$Q_1$ is the heat absorbed in the conversion of ice to water at temperature of $0\text{°C}$.

$m$ is the mass of ice

$L_{\text{f}}$ is the latent heat of fusion

Substitute $1\text{kg}$ for $m$ and $335\text{kJ/kg}$ for $L_{\text{f}}$ to find $Q_1$.

$\begin{array}{c}{Q}_1=\left(1\text{kg}\right)\left(335\text{kJ/kg}\right)\\ =335\text{kJ}\end{array}$

Write the expression for the heat energy liberated when the steam becomes water at the boiling point of water.

$Q_2=m{L}_{\text{v}}$

Here,

$Q_2$ is the heat liberated in the conversion of steam into water at the boiling point of water

$L_{\text{v}}$ is the latent heat of vaporization

Substitute $1\text{kg}$ for $m$ and $2260\text{kJ/kg}$ for $L_{\text{v}}$ to find $Q_2$.

$\begin{array}{c}{Q}_1=\left(1\text{kg}\right)\left(2260\text{kJ/kg}\right)\\ =2260\text{kJ}\end{array}$

Thus the $1\text{kg}$ of steam at $100\text{°C}$ will liberate heat energy of $2260\text{kJ}$ while becoming water at $100\text{°C}$ and $1\text{kg}$ of ice at $0\text{°C}$ will absorb heat energy of $335\text{kJ}$ while becoming water at $0\text{°C}$.

Write the expression for the heat energy gained while increasing the temperature of a substance.

$\begin{array}{c}{Q}_3=mc\Delta T\\ =mc\left(T_2-T_1\right)\end{array}$

Here,

$Q_3$ is the heat energy gained during the temperature increase

$m$ is the mass of the substance

$c$ is the specific heat of the substance

$\Delta T$ is the change in temperature

$T_2$ is the final temperature

$T_1$ is the initial temperature

In the mixing process, the total heat gained by the ice will be equal to the total heat lost by the steam.

$\begin{array}{c}m{L}_{\text{f}}+mc\Delta T_{\text{ice}}=m{L}_{\text{v}}+mc\Delta T_{\text{steam}}\\ m{L}_{\text{f}}+mc\left(T_2-T_{1,\text{ice}}\right)=m{L}_{\text{v}}+mc\left(T_{\text{1,steam}}-T_2\right)\end{array}$

Solve for $T_2$.

$\begin{array}{c}{L}_{\text{f}}+c\left(T_2-T_{1,\text{ice}}\right)=L_{\text{v}}+c\left(T_{\text{1,steam}}-T_2\right)\\ L_{\text{f}}-L_{\text{v}}=c\left(T_{\text{1,steam}}-T_2\right)-c\left(T_2-T_{1,\text{ice}}\right)\\ T_2=\frac{c\left(T_{1,\text{ice}}+T_{\text{1,steam}}\right)+\left(L_{\text{v}}-L_{\text{f}}\right)}{2c}\end{array}$

Substitute $4.186\text{kJ/kg}\cdot °\text{C}$ for $c$, $0\text{°C}$ for $T_{1,\text{ice}}$, $100\text{°C}$ for $T_{\text{1,steam}}$, $335\text{kJ/kg}$ for $L_{\text{f}}$ and $2260\text{kJ/kg}$ for $L_{\text{v}}$ to find the final temperature after mixing $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$.

$\begin{array}{c}{T}_2=\frac{\left[\left(4.186\text{kJ/kg}\cdot °\text{C}\right)\left(0\text{°C}+100\text{°C}\right)\right]+\left[\left(2260\text{kJ/kg}\right)-\left(335\text{kJ/kg}\right)\right]}{2\left(4.186\text{kJ/kg}\cdot °\text{C}\right)}\\ =280\text{°C}>50\text{°C}\end{array}$

Conclusion:

Therefore, the mixing $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$ will result final temperature higher than $50\text{°C}$.

To determine

Whether mixing of given samples will result water having temperature equal to $50\text{°C}$ or a mixture having final temperature less than or greater than $50\text{°C}$.

Answer to Problem 87E

The mixing $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$ will result final temperature lower than $50\text{°C}$.

Explanation of Solution

Given info: The samples are $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$.

In the mixing process, the heat gained by the ice will be equal to the heat lost by the water.

$\begin{array}{c}m{L}_{\text{f}}+mc\Delta T_{\text{ice}}=mc\Delta T_{\text{water}}\\ m{L}_{\text{f}}+mc\left(T_2-T_{1,\text{ice}}\right)=mc\left(T_{\text{1,water}}-T_2\right)\end{array}$

Solve for $T_2$.

$\begin{array}{c}{L}_{\text{f}}+c\left(T_2-T_{1,\text{ice}}\right)=c\left(T_{\text{1,water}}-T_2\right)\\ L_{\text{f}}+2c{T}_2=c\left(T_{1,\text{ice}}+T_{\text{1,water}}\right)\\ T_2=\frac{c\left(T_{1,\text{ice}}+T_{\text{1,water}}\right)-L_{\text{f}}}{2c}\end{array}$

Substitute $4.186\text{kJ/kg}\cdot °\text{C}$ for $c$,$0\text{°C}$ for $T_{1,\text{ice}}$, $100\text{°C}$ for $T_{\text{1,water}}$ and $335\text{kJ/kg}$ for $L_{\text{f}}$ to find the final temperature after mixing $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$.

$\begin{array}{c}{T}_2=\frac{\left[\left(4.186\text{kJ/kg}\cdot °\text{C}\right)\left(0\text{°C}+100\text{°C}\right)\right]-335\text{kJ/kg}}{2\left(4.186\text{kJ/kg}\cdot °\text{C}\right)}\\ =9.98\text{°C}<50\text{°C}\end{array}$

Conclusion:

Therefore, the mixing $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$ will result final temperature lower than $50\text{°C}$.

To determine

Whether mixing of given samples will result water having temperature equal to $50\text{°C}$ or a mixture having final temperature less than or greater than $50\text{°C}$.

Answer to Problem 87E

The mixing $1\text{kg}$ of water at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$ will result final temperature higher than $50\text{°C}$.

Explanation of Solution

Given info: The samples are $1\text{kg}$ of water at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$.

In the mixing process, the heat gained by the water will be equal to the heat lost by the steam.

$\begin{array}{c}mc\Delta T_{\text{water}}=m{L}_{\text{v}}+mc\Delta T_{\text{steam}}\\ mc\left(T_2-T_{1,\text{water}}\right)=m{L}_{\text{v}}+mc\left(T_{\text{1,steam}}-T_2\right)\end{array}$

Solve for $T_2$.

$\begin{array}{c}c\left(T_2-T_{1,\text{water}}\right)=L_{\text{v}}+c\left(T_{\text{1,steam}}-T_2\right)\\ 2c{T}_2-c{T}_{1,\text{water}}=L_{\text{v}}+c{T}_{\text{1,steam}}\\ 2c{T}_2=c\left(T_{1,\text{water}}+T_{\text{1,steam}}\right)+L_{\text{v}}\\ T_2=\frac{c\left(T_{1,\text{water}}+T_{\text{1,steam}}\right)+L_{\text{v}}}{2c}\end{array}$

Substitute $4.186\text{kJ/kg}\cdot °\text{C}$ for $c$, $0\text{°C}$ for $T_{1,\text{water}}$, $100\text{°C}$ for $T_{\text{1,steam}}$ and $2260\text{kJ/kg}$ for $L_{\text{v}}$ to find the final temperature after mixing $1\text{kg}$ of ice at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$.

$\begin{array}{c}{T}_2=\frac{\left[\left(4.186\text{kJ/kg}\cdot °\text{C}\right)\left(0\text{°C}+100\text{°C}\right)\right]+\left(2260\text{kJ/kg}\right)}{2\left(4.186\text{kJ/kg}\cdot °\text{C}\right)}\\ =320\text{°C}>50\text{°C}\end{array}$

Conclusion:

Therefore, the mixing $1\text{kg}$ of water at $0\text{°C}$ and $1\text{kg}$ of steam at $100\text{°C}$ will result final temperature higher than $50\text{°C}$.

To determine

Whether mixing of given samples will result water having temperature equal to $50\text{°C}$ or a mixture having final temperature less than or greater than $50\text{°C}$.

Answer to Problem 87E

Mixing of $1\text{kg}$ of water at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$ will result water having temperature equal to $50\text{°C}$.

Explanation of Solution

Given info: The samples are $1\text{kg}$ of water at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$.

In the mixing process, the heat gained by the water will be equal to the heat lost by the steam.

$\begin{array}{c}mc\Delta T_{\text{water}}=m{L}_{\text{v}}+mc\Delta T_{\text{steam}}\\ mc\left(T_2-T_{\text{1,cold water}}\right)=mc\left(T_{\text{1,hot water}}-T_2\right)\end{array}$

Solve for $T_2$.

$\begin{array}{c}{T}_2-T_{\text{1,cold water}}=T_{\text{1,hot water}}-T_2\\ 2T_2=T_{\text{1,cold water}}+T_{\text{1,hot water}}\\ T_2=\frac{T_{\text{1,cold water}}+T_{\text{1,hot water}}}{2}\end{array}$

Substitute $0\text{°C}$ for $T_{\text{1,cold water}}$ and $100\text{°C}$ for $T_{\text{1,hot water}}$ to find the final temperature after mixing $1\text{kg}$ of water at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$.

$\begin{array}{c}{T}_2=\frac{0\text{°C}+100\text{°C}}{2}\\ =50\text{°C}\end{array}$

Conclusion:

Therefore, mixing of $1\text{kg}$ of water at $0\text{°C}$ and $1\text{kg}$ of water at $100\text{°C}$ will result water having temperature equal to $50\text{°C}$.