Solved Problem on Heat

Determine the heat required to convert 100 g of ice at −10 °C in 100 g of steam at 100 °C. Also, plot a graph of the temperature as a function of the quantity of heat of the phase changes. Data:
specific heat of ice:    c_i = 0.5 cal/g°C;
latent heat of fusion:    L_F = 80 cal/g;
specific heat of water:    c_w = 1.0 cal/g°C;
latent heat of vaporization:    L_v = 540 cal/g.

Problem data:

Mass of ice: m = 100 g;
Initial temperature of ice: t_i = −10 °C;
Final temperature of steam: t_f = 100 °C;
Specific heat of ice: c_i = 0.5 cal/g°C;
Latent heat of fusion: L_F = 80 cal/g;
Specific heat of water: c_w = 1.0 cal/g°C;
Latent heat of vaporization: L_v = 540 cal/g.

Solution

1.ª Phase

First, all ice must be heated from −10 °C to 0 °C (Figure 1), during this phase there is no phase change (the ice does not melt). Using the equation of heat, we have

Figure 1

\[ \begin{gather} \bbox[#99CCFF,10px] {Q=m c \Delta t} \tag{I} \end{gather} \]

\[ \begin{gather} Q_{1}=m c_{g}\Delta t\\ Q_{1}=m c_{g}(t_{0}-t_{-10})\\ Q_{1}=100\times 0.5\times [0-(-10)]\\ Q_{1}=100\times 0.5\times 10\\ Q_{1}=500\ \text{cal} \end{gather} \]

2.ª Phase

The ice must melt from the solid state to the liquid, during this phase the temperature remains at 0 °C (Figure 2). Using the equation for latent heat, we have

Figure 2

\[ \begin{gather} \bbox[#99CCFF,10px] {Q=m L} \tag{I} \end{gather} \]

\[ \begin{gather} Q_{2}=m L_{F}\\ Q_{2}=100\times 80\\ Q_{2}=8000\ \text{cal} \end{gather} \]

3.ª Phase

Water must be heated from 0 °C to 100 °C (Figure 3), during this phase there is no phase change (water does not vaporize), using the expression (I)

Figure 3

\[ \begin{gather} Q_{3}=m c_{a}\Delta t\\ Q_{3}=m c_{a}(t_{100}-t_{0})\\ Q_{3}=100\times 1.0\times (100-0)\\ Q_{3}=100\times 1.0\times 100\\ Q_{3}=10000\ \text{cal} \end{gather} \]

4.ª Phase

Water should evaporate passing from the liquid to the steam, during this phase the temperature remains at 100 °C (Figure 4), using the expression (II)

Figure 4

\[ \begin{gather} Q_{4}=m L_{v}\\ Q_{2}=100\times 540\\ Q_{4}=54000\ \text{cal} \end{gather} \]

Thus the total heat to convert 100 g of ice into −10 °C of steam at 100 °C will be the sum of all parts calculated above.

\[ \begin{gather} Q=Q_{1}+Q_{2}+Q_{3}+Q_{4}\\ Q=500+8000+10000+54000 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {Q=72500\ \text{cal}} \]

Plotting in a graph, the values of the temperatures of each phase of the transformations and the amounts of heat accumulated at each phase we have the graph of Figure 5 below

Figure 5

Note: See that, made the graph on a scale, the amount of heat required to heat the ice of −10 °C to 0 °C is represented by a small part, while the amount of heat required to vaporize the water at 100 °C occupies most of the graph.

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .