Chapter 14, Problem 42P

To determine

The percentage change in answer of problem 41 for glass with mass $350\text{g}$ and specific heat $0.837\text{kJ}/\text{kg}\cdot \text{K}$.

Answer to Problem 42P

The percentage change is $35\%$.

Explanation of Solution

Refer problem 41.

Write the equation for net heat energy of ice-tea system.

$Q_t+Q_{ice}+Q_g=0$ (I)

Here, $Q_t$ is the heat energy possessed by tea, $Q_{ice}$ is the heat energy possessed by ice, and $Q_g$ is the heat energy possessed by glass.

Change in temperature of tea and glass.

Write the equation for $Q_t$.

$Q_t={\rho }_w{V}_t{c}_w\left(T_f-T_i\right)$ (II)

Here, ${\rho }_w$ is the density of water, $V_t$ of tea, $c_w$ is the specific heat of water,$T_i$ is the initial temperature of tea and glass, and $T_f$ is the final temperature of tea and glass.

Write the equation for $Q_{ice}$.

$Q_{ice}=m_{ice}{L}_f+m_{ice}{c}_{ice}\left(T_{1f}-T_{1i}\right)+m_{ice}{c}_w\left(T_{2f}-T_{2i}\right)$ (III)

Here, $m_{ice}$ is the mass of ice, $L_f$ is the latent heat of fusion of ice, $c_{ice}$ is the specific heat of ice, $T_{1i}$ is the initial temperature of ice, $T_{1f}$ is the final temperature of ice, temperature change of ice,$T_{2f}$ is the final temperature of ice-water, and $T_{2i}$ is the final temperature of ice-water.

Write the equation for $Q_g$.

$Q_g=m_g{c}_g\left(T_f-T_i\right)$ (IV)

Here, $m_g$ is the mass of glass and $c_g$ is the specific heat of glass.

Rewrite equation (I) by substituting equations (II), (III), and (IV).

${\rho }_w{V}_t{c}_w\left(T_f-T_i\right)+\left[m_{ice}{L}_f+m_{ice}{c}_{ice}\left(T_{1f}-T_{1i}\right)+m_{ice}{c}_w\left(T_{2f}-T_{2i}\right)\right]+m_g{c}_g\left(T_f-T_i\right)=0$

Rewrite the above relation in terms of $m_{ice}$.

$m_{ice}=\frac{-\left({\rho }_w{V}_t{c}_w+m_g{c}_g\right)\left(T_f-T_i\right)}{L_f+c_{ice}\left(T_{1f}-T_{1i}\right)+c_w\left(T_{2f}-T_{2i}\right)}$

Write the equation to find the percentage change.

$\Delta =\left(\frac{m_{ice}-m_i}{m_i}\right)100\%$

Here,$\Delta$ is the percentage change and $m_i$ is the mass of ice calculated in problem 41.

Conclusion:

Substitute $1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_w$, $2.00\times {10}^{-4}{\text{m}}^{\text{3}}$ for $V_t$, $4.186\text{kJ}/\text{kg}\cdot \text{K}$ for $c_w$, $0.35\text{kg}$ for $m_g$, $0.837\text{kJ}/\text{kg}\cdot \text{K}$ for $c_g$, $10.0°\text{C}$ for $T_f$, $95.0°\text{C}$ for $T_i$, $333.7\text{kJ}/\text{kg}$ for $L_f$, $2.1\text{kJ}/\text{kg}\cdot \text{K}$ for $c_{ice}$, $-10.0°\text{C}$ for $T_{1i}$ , $0°\text{C}$ for $T_{1f}$, $0°\text{C}$ for $T_{2i}$, and $10.0°\text{C}$ for $T_{2f}$ in the equation for $m_{ice}$.

$\begin{array}{c}{m}_{ice}=\frac{-\left(\begin{array}{l}\left(\left(1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(2.00\times {10}^{-4}{\text{m}}^{\text{3}}\right)\left(4.186\text{kJ}/\text{kg}\cdot \text{K}\right)\right)+\\ \left(\left(0.35\text{kg}\right)\left(0.837\text{kJ}/\text{kg}\cdot \text{K}\right)\right)\end{array}\right)\left(10°\text{C}-\left(95.0°\text{C}\right)\right)}{\left(\begin{array}{l}333.7\text{kJ}/\text{kg}+\left(\left(2.1\text{kJ}/\text{kg}\cdot \text{K}\right)\left(0°\text{C}-\left(-10.0°\text{C}\right)\right)\right)+\\ \left(4.186\text{kJ}/\text{kg}\cdot \text{K}\right)\left(10.0°\text{C}-\left(0°\text{C}\right)\right)\end{array}\right)}\\ =\frac{-\left(\begin{array}{l}\left(\left(1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(2.00\times {10}^{-4}{\text{m}}^{\text{3}}\right)\left(4.186\text{kJ}/\text{kg}\cdot \text{K}\right)\right)+\\ \left(\left(0.35\text{kg}\right)\left(0.837\text{kJ}/\text{kg}\cdot \text{K}\right)\right)\end{array}\right)\left(\left(10+273\right)\text{K}-\left(95.0+273\right)\text{K}\right)}{\left(\begin{array}{l}333.7\text{kJ}/\text{kg}+\left(\left(2.1\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(0+273\right)\text{K}-\left(-10.0+273\right)\text{K}\right]\right)+\\ \left(4.186\text{kJ}/\text{kg}\cdot \text{K}\right)\left[\left(10.0+273\right)\text{K}-\left(0+273\right)\text{K}\right]\end{array}\right)}\\ =\frac{\left(837.2\text{J/K}+292.95\text{J/K}\right)\left(85\text{K}\right)}{\left(354.7\text{kJ}/\text{kg}+41.86\text{kJ/kg}\right)}\\ =0.242\text{kg}\left(\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1}\text{kg}}\right)\\ =242\text{g}\end{array}$.

Substitute $242\text{g}$ for $m_{ice}$ and $179\text{g}$ for $m_i$ in the equation for $\Delta$.

$\begin{array}{c}\Delta =\left(\frac{242\text{g}-179\text{g}}{179\text{g}}\right)100\%\\ =35\%\end{array}$

Therefore, the percentage change is $35\%$.