Question

(II) What mass of steam at 100°C must be added to 1.00 kg of ice at 0°C to yield liquid water at 30°C?

Short answer

The amount of steam added to the ice to yield liquid water is 0.18 kg.

Step-by-step solution

Understanding of conservation of heat energy

When different parts of an isolated system are at different temperatures, heat energy flows from the part which is at a higher temperature to the part which is at a lower temperature until thermal equilibrium is reached.

Thus, according to the law of conservation of energy in an isolated system, the heat lost by one part of the system must be equal to the heat gained by the other part.

Heat lost = heat gained

Given information

The temperature of steam is \({T_1} = 100\circ {\rm{C}}\).

Mass of ice is \(m = 1.00\;{\rm{kg}}\).

The temperature of ice is \({T_2} = 0\circ {\rm{C}}\).

The temperature of liquid water is \(T = 30\circ {\rm{C}}\).

Specific heat of water is \(c = 4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}\).

Latent heat of vaporization is \({L_{\rm{V}}} = 22.6 \times {105}\;{\rm{J/kg}}\).

Latent heat of fusion is \({L_{\rm{F}}} = 3.33 \times {105}\;{\rm{J/kg}}\).

Determination of the mass of the steam added to yield water

Let \(m'\)be the mass of the steam that must be added to the ice.

The steam at\(100\circ {\rm{C}}\)is first converted to water at\(100\circ {\rm{C}}\)and then its temperature changes from\(100\circ {\rm{C}}\)to\(30\circ {\rm{C}}\).

The total heat lost in this process is given by

\({Q_{{\rm{Lost}}}} = m'{L_{\rm{v}}} + m'c\left( {{T_1} - T} \right)\).

Ice at\(0\circ {\rm{C}}\)is first converted into the water at\(0\circ {\rm{C}}\), and then its temperature rises to\(30\circ {\rm{C}}\).

The total heat gained in this process is given by

\({Q_{{\rm{gained}}}} = m{L_{\rm{F}}} + mc\left( {T - {T_2}} \right)\).

By applying the law of conservation of energy, you get the following equation:

\(\begin{array}{c}{\rm{Heat}}\;{\rm{lost}}\;{\rm{by steam}} = {\rm{Heat}}\;{\rm{gained}}\;{\rm{by ice}}\\m'{L_{\rm{v}}} + m'c\left( {{T_1} - T} \right) = m{L_{\rm{F}}} + mc\left( {T - {T_2}} \right)\\m' = \frac{{m\left\{ {{L_{\rm{F}}} + c\left( {T - {T_2}} \right)} \right\}}}{{{L_{\rm{V}}} + c\left( {{T_1} - T} \right)}}\end{array}\)

Substitute the values into the above equation.

\(\begin{array}{c}m' = \frac{{\left( {1.00\;{\rm{kg}}} \right)\left\{ {\left( {3.33 \times {{10}5}\;{\rm{J/kg}}} \right){\rm{ + }}\left( {4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}} \right)\left( {30\circ {\rm{C}}} \right)} \right\}}}{{\left( {22.6 \times {{10}5}\;{\rm{J/kg}}} \right){\rm{ + }}\left( {4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}} \right)\left( {70\circ {\rm{C}}} \right)}}\\ = \frac{{458580}}{{2553020}}\\ = 0.18\;{\rm{kg}}\end{array}\)

Thus, the amount of steam added to the ice to yield liquid water is 0.18 kg.