Chapter 14, Problem 110P

To determine

Find the final temperature of the water.

Answer to Problem 110P

The final temperature of the water is $22.6°\text{C}$.

Explanation of Solution

Write the equation for heat energy required to produce temperature change in water.

$Q_{\text{W}}=m_{\text{W}}{c}_{\text{W}}\left(T_{\text{f}}-T_{\text{i}}\right)$ (I)

Here, $Q_{\text{W}}$ is the heat energy acquired by the water, $m_{\text{W}}$ is the mass of the water, $c_{\text{W}}$ is the specific heat capacity of water, $T_{\text{f}}$ is the final temperature of the system, and $T_{\text{i}}$ is the initial temperature of the system.

Write the equation for heat energy required to produce temperature change in gold.

$Q_{\text{Au}}=m_{\text{Au}}{c}_{\text{Au}}\left(T_{\text{f}}-T_{\text{Au}}\right)$ (II)

Here, $Q_{\text{Fe}}$ is the heat energy acquired by the gold, $m_{\text{Au}}$ is the mass of the gold, $c_{\text{Au}}$ is the specific heat capacity of the gold, and $T_{\text{Au}}$ is the initial temperature of the gold.

Write the equation for heat energy required to produce temperature change in copper pot

$Q_{\text{Cu}}=m_{\text{Cu}}{c}_{\text{Cu}}\left(T_{\text{f}}-T_{\text{i}}\right)$ (III)

Here, $Q_{\text{Cu}}$ is the heat energy acquired by the copper, $m_{\text{Cu}}$ is the mass of the copper, and $c_{\text{Cu}}$ is the specific heat capacity of the copper.

According to the thermal equilibrium condition, heat flows from the gold to the water and copper pot is equal to zero.

$Q_{\text{W}}+Q_{\text{Au}}+Q_{\text{Cu}}=0$ (IV)

Conclusion:

Substitute the equation (I), (II), and (III) in equation (IV).

$\begin{array}{c}{m}_{\text{W}}{c}_{\text{W}}\left(T_{\text{f}}-T_{\text{i}}\right)+m_{\text{Au}}{c}_{\text{Au}}\left(T_{\text{f}}-T_{\text{Au}}\right)+m_{\text{Cu}}{c}_{\text{Cu}}\left(T_{\text{f}}-T_{\text{i}}\right)=0\\ \left(m_{\text{W}}{c}_{\text{W}}\right)T_{\text{f}}-\left(m_{\text{W}}{c}_{\text{W}}\right)T_{\text{i}}+\left(m_{\text{Au}}{c}_{\text{Au}}\right)T_{\text{f}}-\left(m_{\text{Au}}{c}_{\text{Au}}\right)T_{\text{Au}}+\left(m_{\text{Cu}}{c}_{\text{Cu}}\right)T_{\text{f}}-\left(m_{\text{Cu}}{c}_{\text{Cu}}\right)T_{\text{i}}=0\\ \left(m_{\text{W}}{c}_{\text{W}}+m_{\text{Au}}{c}_{\text{Au}}+m_{\text{Cu}}{c}_{\text{Cu}}\right)T_{\text{f}}-\left(m_{\text{W}}{c}_{\text{W}}+m_{\text{Cu}}{c}_{\text{Cu}}\right)T_{\text{i}}-\left(m_{\text{Au}}{c}_{\text{Au}}\right)T_{\text{Au}}=0\end{array}$

Solve the above equation for final temperature.

$\begin{array}{c}\left(m_{\text{W}}{c}_{\text{W}}+m_{\text{Au}}{c}_{\text{Au}}+m_{\text{Cu}}{c}_{\text{Cu}}\right)T_{\text{f}}=\left(m_{\text{W}}{c}_{\text{W}}+m_{\text{Cu}}{c}_{\text{Cu}}\right)T_{\text{i}}+\left(m_{\text{Au}}{c}_{\text{Au}}\right)T_{\text{Au}}\\ T_{\text{f}}=\frac{\left(m_{\text{W}}{c}_{\text{W}}+m_{\text{Cu}}{c}_{\text{Cu}}\right)T_{\text{i}}+\left(m_{\text{Au}}{c}_{\text{Au}}\right)T_{\text{Au}}}{\left(m_{\text{W}}{c}_{\text{W}}+m_{\text{Au}}{c}_{\text{Au}}+m_{\text{Cu}}{c}_{\text{Cu}}\right)}\end{array}$

Substitute $0.500\text{L}$ for $m_{\text{W}}$, $4186\text{J/kg}\cdot \text{K}$ for $c_{\text{W}}$, $1.500\text{kg}$ for $m_{\text{Cu}}$, $385\text{J/kg}\cdot \text{K}$ for $c_{\text{Cu}}$, $22.0°\text{C}$ for $T_{\text{i}}$, $0.250\text{kg}$ for $m_{\text{Au}}$, $128\text{J/kg}\cdot \text{K}$ for $c_{\text{Au}}$, and $75.0°\text{C}$ for $T_{\text{Au}}$ in above equation to find $T_{\text{f}}$.

$\begin{array}{c}{T}_{\text{f}}=\frac{\left[\left(0.500\text{L}\right)\left(\frac{1\text{kg}}{1\text{L}}\right)\left(4186\text{J/kg}\cdot \text{K}\right)+\left(1.500\text{kg}\right)\left(385\text{J/kg}\cdot \text{K}\right)\right]22.0\text{K}+\left[\left(0.250\text{kg}\right)\left(128\text{J/kg}\cdot \text{K}\right)\right]75.0\text{K}}{\left[\left(0.500\text{L}\right)\left(\frac{1\text{kg}}{1\text{L}}\right)\left(4186\text{J/kg}\cdot \text{K}\right)+\left(1.500\text{kg}\right)\left(385\text{J/kg}\cdot \text{K}\right)+\left(0.250\text{kg}\right)\left(128\text{J/kg}\cdot \text{K}\right)\right]}\\ =\frac{\left(2093\text{J/K}+577.5\text{J/K}\right)22.0\text{K}+\left(32\text{J/K}\right)75.0\text{K}}{\left(2702.5\text{J/K}\right)}\\ =22.6°\text{C}\end{array}$

Therefore, the final temperature of the water is $22.6°\text{C}$.