Chapter 18, Problem 37SP

Determine the result when 100 g of steam at 100° C is passed into a mixture of 200 g of water and 20 g of ice at exactly 0 °C in a calorimeter that behaves thermally as if it were equivalent to 30 g of water.

To determine

The result when $100\text{ g}$ of steam at $100°\text{C}$ is passed through a mixture of $200\text{ g}$ of water and $20\text{ g}$ of ice, both at $0°\text{C}$.

Answer to Problem 37SP

Solution:

$49\text{ g}$ of steam is condensed, final temperature is $100°\text{C}$.

Explanation of Solution

Given data:

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

The mass of water is $200\text{ g}$.

The mass of ice is $20\text{ g}$.

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 $30\text{ g}$ 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 less than $100°\text{C}$, that is, all the steam condenses 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 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 $100\text{ g}$ for $m_1$ and $540\text{ cal/g}$ for $L_{v1}$.

$\begin{array}{c}{Q}_{v1}=\left(100\text{ g}\right)\left(540\text{ cal/g}\right)\\ =54000\text{ cal}\end{array}$

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

Write the expression for the change in temperature of 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 $100\text{ g}$ for $m_1$

$\begin{array}{c}{Q}_1=\left(100\text{ g}\right)\left[1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(100°\text{C}-T_1\right)\\ =10000-100T_1\end{array}$

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 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 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 $\text{20 g}$ for $m_i$ and $80\text{ cal/g}$ for $L_{f1}$

$\begin{array}{c}{Q}_{f1}=\left(\text{20 g}\right)\left(80\text{ cal/g}\right)\\ =1600\text{ cal}\end{array}$

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 form 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 $\text{20 g}$ for $m_i$

$\begin{array}{c}{Q}_3=\left(20\text{ g}\right)\left[1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(T_1\right)\\ =20T_1\end{array}$

Recall the expression for heat transfer due to temperature difference in order to calculate the heat removed from $200\text{ g}$ 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 $200\text{ g}$ for $m_w$

$\begin{array}{c}{Q}_2=\left(200\text{ g}\right)\left[1\text{ cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(T_1\right)\\ =200T_1\end{array}$

Understand that the calorimeter is equivalent to $30\text{ g}$ 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 $30\text{ g}$ for $m_2$

$\begin{array}{c}{Q}_4=\left(30\text{ g}\right)\left[\text{1 cal/}\left(\text{g}\cdot °\text{C}\right)\right]\left(T_1\right)\\ =30T_1\end{array}$

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

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

Substitute $200T_1$ for $Q_2$, $30T_1$ for $Q_4$, $20T_1$ for $Q_3$, $10000-100T_1$ for $Q_1$, $54000\text{ cal}$ for $Q_{v1}$, and $1600\text{ cal}$ for $Q_{f1}$

$\begin{array}{c}54000\text{ cal}+\left(10000-100T_1\right)-\left(1600\text{ cal}\right)-\left(200T_1\right)-\left(20T_1\right)-\left(30T_1\right)=0\\ 54000+10000-1600=100T_1+200T_1+20T_1+30T_1\\ 62400=350T_1\\ T_1=\frac{62400}{350}\end{array}$

Further solve as,

$T_1\simeq 178°\text{C}$

Understand that the final temperature of the mixture cannot be more than $100°\text{C}$. This indicates that our assumption that the final temperature of the mixture is less than $100°\text{C}$ is incorrect. Therefore, not all the steam condenses. Hence, the final temperature of the mixture is $100°\text{C}$.

Rewrite the expression for heat released by the mass of steam condensed.

$Q_{v2}=m_s{L}_{v1}$

Here, $Q_{v2}$ is the heat released by the mass of steam condensed and $m_s$ is the mass of steam condensed.

Substitute $540\text{ g}$ for $L_{v1}$

$\begin{array}{c}{Q}_{v2}=m_s\left(540\text{ cal/g}\right)\\ =540m_s\end{array}$

The final temperature of the mixture is $100°\text{C}$. This indicates that the heats $Q_1$ and $Q_{v1}$ will not exist in the final calorimeter heat expression. Rewrite the expression for heat transferred in calorimeter.

$\begin{array}{c}{\displaystyle \sum Q_{net}}=0\\ Q_{v2}-Q_{f1}-Q_2-Q_3-Q_4=0\end{array}$

Substitute $200T_1$ for $Q_2$, $30T_1$ for $Q_4$, $20T_1$ for $Q_3$, $540m_s$ for $Q_{v2}$, and $1600\text{ cal}$ for $Q_{f1}$

$\begin{array}{c}540m_s-\left(1600\text{ cal}\right)-\left(200T_1\right)-\left(20T_1\right)-\left(30T_1\right)=0\\ 540m_s-1600=200T_1+20T_1+30T_1\\ 540m_s=250T_1+1600\end{array}$

The final temperature of the mixture is $100°\text{C}$. Therefore, substitute $100°\text{C}$ for $T_1$.

$\begin{array}{c}540m_s=250\left(100°\text{C}\right)+1600\\ 540m_s=25000+1600\\ 540m_s=26600\\ m_s=\frac{26600}{540}\end{array}$

Further solve as,

$m_s\simeq 49\text{ g}$

Conclusion:

The result is that $49\text{ g}$ of steam is condensed and the final temperature of the mixture is $100°\text{C}$.