Chapter 5, Problem 35MC

To determine

The result of losing $3\text{MJ}$ heat from $1\text{kg}$ of steam at $200\text{°C}$.

Answer to Problem 35MC

Option (b) the result of losing $3\text{MJ}$ heat from $1\text{kg}$ of steam at $200\text{°C}$ is water and ice.

Explanation of Solution

Write the expression for the expression for the heat liberated when temperature of a substance reduces.

$\begin{array}{c}Q=mc\Delta T\\ =mc\left(T_1-T_2\right)\end{array}$ (1)

Here,

$Q$ is the amount of heat added to the system

$m$ is the mass of substance

$c$ is the specific heat of substance

$\Delta T$ is the change in temperature

$T_2$ is the final temperature

$T_1$ is the initial temperature

The specific heat of steam is $1.996\text{kJ/kg}\cdot \text{°C}$.

Substitute $1\text{kg}$ for $m$, $1.996\text{kJ/kg}\cdot \text{°C}$ for $c$, $200\text{°C}$ for $T_1$ and $100\text{°C}$ for $T_2$ to find the heat liberated when $1\text{kg}$ of steam at $200\text{°C}$ becomes steam at $100\text{°C}$.

$\begin{array}{c}Q=\left(1\text{kg}\right)\left(1.996\text{kJ/kg}\cdot \text{°C}\right)\left(200\text{°C}-100\text{°C}\right)\\ =199\text{kJ}\end{array}$

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

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

Here,

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

The heat of vaporization for water is $2260\text{kJ/kg}$.

Substitute $1\text{kg}$ for $m$ and $2260\text{kJ/kg}$ for $L_{\text{v}}$ to find the heat liberated when $1\text{kg}$ of steam at $100\text{°C}$ becomes water at $100\text{°C}$.

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

The specific heat of water is $4.186\text{kJ/kg}\cdot \text{°C}$.

Substitute $1\text{kg}$ for $m$, $4.186\text{kJ/kg}\cdot \text{°C}$ for $c$, $100\text{°C}$ for $T_1$ and $0\text{°C}$ for $T_2$ in equation (1) to find the heat energy liberated when $1\text{kg}$ of water at $100\text{°C}$ becomes water at $0\text{°C}$.

$\begin{array}{c}Q=\left(1\text{kg}\right)\left(4.186\text{kJ/kg}\cdot \text{°C}\right)\left(100\text{°C}-0\text{°C}\right)\\ =418.6\text{kJ}\end{array}$

Thus, the total heat liberated when $1\text{kg}$ of steam at $200\text{°C}$ becomes water at $0\text{°C}$ is,

$\begin{array}{c}{Q}_{\text{total}}=199\text{kJ}+2260\text{kJ}+418.6\text{kJ}\\ =2878\text{kJ}\\ =2878\text{kJ}\times \frac{{10}^{-3}\text{MJ}}{1\text{kJ}}\\ =2.9\text{MJ}\end{array}$

When the steam liberate $2.9\text{MJ}$ heat, it will completely turn into water at $0\text{°C}$. Sill there is $0.1\text{MJ}$ of energy remaining in $3\text{MJ}$.

Write the expression for the heat corresponding to the phase change of water at $0\text{°C}$ to ice at $0\text{°C}$.

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

Here,

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

The heat of fusion of water is $335\text{kJ/kg}$.

Substitute $0.1\text{MJ}$ for $Q$ and $335\text{kJ/kg}$ for $L_{\text{f}}$ to find the amount of water at $0\text{°C}$ becomes ice at $0\text{°C}$ by liberating heat equal to $0.1\text{MJ}$.

$\begin{array}{c}0.1\text{MJ}=m\left(335\text{kJ/kg}\right)\\ m=\frac{0.1\text{MJ}}{335\text{kJ/kg}}\\ =\frac{0.1\text{MJ}\times \frac{{10}^3\text{kJ}}{1\text{MJ}}}{335\text{kJ/kg}}\\ =0.3\text{kg}\end{array}$

So, liberation the $0.1\text{MJ}$ heat will cause $0.3\text{kg}$ of water at $0\text{°C}$ to turn into ice at $0\text{°C}$. Thus, the result of losing $3\text{MJ}$ heat from $1\text{kg}$ of steam at $200\text{°C}$ will be a mixture of water and ice.

Conclusion:

Since, the liberated heat energy of $3\text{MJ}$ is enough to result the $1\text{kg}$ steam at $200\text{°C}$ to become water and ice, the final state of the system will be a mixture of water and ice. Thus, option (b) is correct.

The liberated heat energy of $3\text{MJ}$ is not enough to result the whole $1\text{kg}$ steam at $200\text{°C}$ to become ice. Thus, option (a) is incorrect.

Since there is energy lost from the system even after it become completely water, the final state is not only consist of water but also ice. Thus, option (c) is incorrect.

Losing $3\text{MJ}$ heat from $1\text{kg}$ of steam at $200\text{°C}$ will definitely convert all the steam into other forms. Thus, option (d) is incorrect.