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 temperature graph 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 Part

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

Figure 1

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

\[ \begin{gather} Q_1=mc_i\Delta t \\[5pt] Q_1=m c_i(t_0-t_{-10}) \\[5pt] Q_1=(100\;\mathrm{\cancel g})\left(0.5\;\mathrm{\small{\frac{cal}{\cancel g\cancel{°C}}}}\right)[0-(-10)]\;\mathrm{\cancel{°C}} \\[5pt] Q_1=100\times(0.5\;\mathrm{cal})\times 10 \\[5pt] Q_1=500\;\mathrm{cal} \end{gather} \]

2.ª Part

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

Figure 2

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

\[ \begin{gather} Q_2=mL_F \\[5pt] Q_2=100\;\mathrm{\cancel g}\times 80\;\mathrm{\small{\frac{cal}{\cancel g}}} \\[5pt] Q_2=8000\;\mathrm{cal} \end{gather} \]

3.ª Part

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

Figure 3

\[ \begin{gather} Q_3=mc_w\Delta t \\[5pt] Q_3=m c_w(t_{100}-t_0) \\[5pt] Q_3=(100\;\mathrm{\cancel g})\left(1.0\;\mathrm{\small{\frac{cal}{\cancel g\cancel{°C}}}}\right)(100-0)\;\mathrm{\cancel{°C}} \\[5pt] Q_3=100\times(1.0\;\mathrm{cal})\times 100 \\[5pt] Q_3=10000\;\mathrm{cal} \end{gather} \]

4.ª Part

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

Figure 4

\[ \begin{gather} Q_4=mL_v \\[5pt] Q_4=100\;\mathrm{\cancel g}\times 540\;\mathrm{\small{\frac{cal}{\cancel g}}} \\[5pt] Q_4=54000\;\mathrm{cal} \end{gather} \]

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.

\[ \begin{gather} Q=Q_1+Q_2+Q_3+Q_4 \\[5pt] Q=(500+8000+10000+54000)\;\mathrm{cal} \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {Q=72500\;\mathrm{cal}} \end{gather} \]

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: if the graph is plotted to scale, the amount of heat required to heat the ice from −10 °C to 0 °C is represented by a very small section (highlighted), while the amount of heat needed 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 .