Chapter 18, Problem 38SP

Determine the result when 10 g of steam at 100 °C is passed into a mixture of 400 g of water and 100 g of ice at exactly 0 °C in a calorimeter that behaves thermally as if it were equivalent to 50 g of water.

To determine

The result when $10\text{ g}$ of steam at $100°\text{C}$ is passed through a mixture of $400\text{ g}$ of water and $100\text{ g}$ of ice both at $0°\text{C}$.

Answer to Problem 38SP

Solution:

$80\text{ g}$ of ice melts, final temperature $0°\text{C}$.

Explanation of Solution

Given data:

The mass of steam is $10\text{ g}$.

The mass of water is $400\text{ g}$.

The mass of ice is $100\text{ g}$.

The initial temperature of water and ice is $0°\text{C}$.

The initial temperature of steam is $100°\text{C}$.

The calorimeter behaves thermally equivalent to $50\text{ g}$ of water.

Formula used:

The heat interaction of a body due to change in temperature is expressed by the formula,

$\Delta Q=mc\Delta T$

Here, $\Delta Q$ is the heat transferred to the body, $\Delta T$ is the change in temperature of the body, $m$ is the mass of the body, and $c$ is the specific heat of the body.

The expression for heat of fusion is,

$Q_f=m{L}_f$

Here, $L_f$ is the latent heat of fusion of the body and $Q_f$ is the heat of fusion.

The expression for heat of condensation is,

$Q_v=m{L}_v$

Here, $m$ is the mass of steam, $L_v$ is the latent heat of condensation of body, and $Q_v$ is the heat required to condense mass $m$.

The expression for net heat transfer in a calorimeter is written as,

${\displaystyle \sum Q_{net}}=0$

Here, ${\displaystyle \sum Q_{net}}$ is the net heat transferred in a calorimeter.

Explanation:

Consider that the final temperature of the mixture is more than $0°\text{C}$, that is, all the ice melts into water and the heat that is released by the steam is gained by the water-and-ice mixture initially at $0°\text{C}$.

Calculate the heat gained by the steam to condense into water.

$Q_{v1}=m_1{L}_{v1}$

Here, $m_1$ is the mass of steam, $L_{v1}$ is the latent heat of condensation of steam, and $Q_{v1}$ is heat released by $100°\text{C}$ steam of mass $m_1$ to condense into water at $100°\text{C}$.

The standard value of latent heat of condensation of steam is $540\text{ cal/g}$. Therefore, substitute $10\text{ g}$ for $m_1$ and $540\text{ cal/g}$ for $L_{v1}$

$\begin{array}{c}{Q}_{v1}=\left(10\text{ g}\right)\left(540\text{ cal/g}\right)\\ =5400\text{ cal}\end{array}$

Understand that after the condensation of all the steam into water at $100°\text{C}$, water loses heat, which leads to a fall in the temperature of water to attain the final temperature of the mixture.

Write the expression for change in temperature of the condensed steam into water from $100°\text{C}$ to the final temperature of the mixture.

$\Delta T_1=100°\text{C}-T_1$

Here, $T_1$ is the final temperature of the mixture, and $\Delta T_1$ is the change in temperature of the steam after condensing into water.

Write the expression for heat lost by water due to change in temperature.

$Q_1=m_1{c}_1\Delta T_1$

Here, $Q_1$ is the heat lost by condensed water, and $c_1$ is the specific heat of water.

The standard value of specific heat of water is $1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$. Substitute $100°\text{C}-T_1$ for $\Delta T_1$, $1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_1$, and $10\text{ g}$ for $m_1$

$\begin{array}{c}{Q}_1=\left(10\text{ g}\right)\left[1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(100°\text{C}-T_1\right)\\ =1000-10T_1\end{array}$

Calculate the change in temperature of ice and water from $0°\text{C}$ to the final temperature of the mixture.

$\begin{array}{c}\Delta T_2=T{-}_{1}0°\text{C}\\ =T_1\end{array}$

Here, $\Delta T_2$ is the change in temperature of ice and water from $0°\text{C}$ to the final temperature of the mixture.

Understand that the change in temperature of the calorimeter is the same as the change in temperature of water and ice.

$\Delta T_c=\Delta T_2$

Here, $\Delta T_c$ is the change in temperature of the calorimeter.

Substitute $T_1$ for $\Delta T_2$

$\Delta T_c=T_1$

Write the expression for the heat required to convert ice at $0°\text{C}$ into water.

$Q_{f1}=m_i{L}_{f1}$

Here, $m_i$ is the mass of ice, $Q_{f1}$ is the heat required to convert ice at $0°\text{C}$ into water, and $L_{f1}$ is the latent heat of fusion of ice at $0°\text{C}$.

The standard value of latent heat of fusion of ice is $80\text{ cal/g}$. Substitute $\text{100 g}$ for $m_i$ and $80\text{ cal/g}$ for $L_{f1}$

$\begin{array}{c}{Q}_{f1}=\left(\text{100 g}\right)\left(80\text{ cal/g}\right)\\ =8000\text{ cal}\end{array}$

Recall the expression for heat transfer due to temperature difference in order to calculate the heat supplied to change the temperature of water at $0°\text{C}$ formed by the melting of ice to water at the final temperature of the mixture.

$Q_3=m_i{c}_1\Delta T_2$

Here, $Q_3$ is the heat supplied to convert $0°\text{C}$ water to the final temperature of the mixture.

Substitute $T_1$ for $\Delta T_2$, $1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_1$, and $\text{100 g}$ for $m_i$

$\begin{array}{c}{Q}_3=\left(\text{100 g}\right)\left[1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(T_1\right)\\ =100T_1\end{array}$

Recall the expression for heat transfer due to temperature difference in order to calculate the heat removed from $400\text{ g}$ of $0°\text{C}$ water to obtain the final temperature of water in the mixture.

$Q_2=m_w{c}_1\Delta T_2$

Here, $m_w$ is the mass of water initially at $0°\text{C}$ and $Q_2$ is the heat removed from the $0°\text{C}$ water to obtain the final temperature of water.

Substitute $-T_1$ for $\Delta T_2$, $1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_1$, and $400\text{ g}$ for $m_w$

$\begin{array}{c}{Q}_2=\left(400\text{ g}\right)\left[1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(T_1\right)\\ =400T_1\end{array}$

Understand that the calorimeter is equivalent to $50\text{ g}$ of water. Recall the expression for heat transfer due to temperature difference in order to calculate the heat removed from the calorimeter at $0°\text{C}$ to obtain the final temperature of the mixture.

$Q_4=m_2{c}_1\Delta T_c$

Here, $m_2$ is the mass of the calorimeter and $Q_4$ is the heat removed from the calorimeter to obtain the final temperature of the mixture.

Substitute $T_1$ for $\Delta T_c$, $1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)$ for $c_1$, and $50\text{ g}$ for $m_2$

$\begin{array}{c}{Q}_4=\left(50\text{ g}\right)\left[\text{1 cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(T_1\right)\\ =50T_1\end{array}$

Recall the expression for the net heat transferred in a calorimeter. Take the heat absorbed by a body as negative and the heat released by a body as positive.

$\begin{array}{c}{\displaystyle \sum Q_{net}}=0\\ Q_{v1}+Q_1-Q_{f1}-Q_2-Q_3-Q_4=0\end{array}$

Substitute $400T_1$ for $Q_2$, $50T_1$ for $Q_4$, $100T_1$ for $Q_3$, $1000-10T_1$ for $Q_1$, $5400\text{ cal}$ for $Q_{v1}$, and $-8000\text{ cal}$ for $Q_{f1}$

$\begin{array}{c}5400\text{ cal}+\left(1000-10T_1\right)-\left(8000\text{ cal}\right)-\left(400T_1\right)-\left(100T_1\right)-\left(50T_1\right)=0\\ 5400+1000-8000=10T_1+400T_1+100T_1+50T_1\\ -1600=560T_1\\ T_1=\frac{-1600}{560}\end{array}$

Further solve as,

$T_1\simeq -2.86°\text{C}$

Understand that the final temperature of the mixture cannot be less than $0°\text{C}$. This indicates that our assumption that the final temperature of mixture is more than $0°\text{C}$ is incorrect. Therefore, not all the ice melts. Hence, the final temperature of the mixture is $0°\text{C}$.

Rewrite the expression for the heat absorbed by the mass of ice melted.

$Q_{f2}=m_{i1}{L}_{f1}$

Here, $Q_{f2}$ is the heat absorbed by the mass of ice melted and $m_{i1}$ is the mass of ice melted.

Substitute $80\text{ cal/g}$ for $L_{v1}$.

$\begin{array}{c}{Q}_{f2}=m_{i1}\left(80\text{ cal/g}\right)\\ =80m_{i1}\end{array}$

The final temperature of the mixture is $0°\text{C}$. This indicates that the heats $Q_{f1}$ and $Q_3$ will not exist in the final calorimeter heat expression. Rewrite the expression for heat transferred in calorimeter.

$\begin{array}{c}{\displaystyle \sum Q_{net}}=0\\ Q_{v1}+Q_1-Q_{f2}-Q_2-Q_4=0\end{array}$

Substitute $200T_1$ for $Q_2$, $30T_1$ for $Q_4$, $5400\text{ cal}$ for $Q_{v2}$, and $80m_{i1}$ for $Q_{f2}$

$\begin{array}{c}5400\text{ cal}+\left(1000-10T_1\right)-\left(80m_{i1}\right)-\left(400T_1\right)-\left(50T_1\right)=0\\ 5400+1000=80m_{i1}+10T_1+400T_1+100T_1+50T_1\\ 5400+1000=80m_{i1}+560T_1\end{array}$

The final temperature of the mixture is $0°\text{C}$. Therefore, substitute $0°\text{C}$ for $T_1$.

$\begin{array}{c}6400=80m_{i1}+560\left(0°\text{C}\right)\\ 6400=80m_{i1}+0\\ 80m_{i1}=6400\\ m_{i1}=\frac{6400}{80}\end{array}$

Further solve as,

$m_{i1}=80\text{ g}$

Conclusion:

Therefore, $80\text{ g}$ of ice melts and the final temperature of the mixture is $0°\text{C}$.