Solved Problem on Heat

In a container thermally insulated from the environment a mixture of ice and water is placed at 0 °C at atmospheric pressure. By supplying a certain amount of energy by heat to the mixture, we verify that the temperature of the mixture does not change and the volume of the system decreases by 0.5 cm³.
a) Calculate the mass of ice that turns into liquid water;
b) Determine the amount of energy by heat received in the mixture.
Data: density of ice 0.92 g/cm³, the density of water 1 g/cm³, and latent heat of fusion of ice 80 cal/g.

Problem data:

Temperature of the water and ice mixture: t = 0 °C;
Mixture volume variation: ΔV = −0.5 cm³;
Water density: d_w = 1 g/cm³;
Ice density: d_i = 0.92 g/cm³;
Latent heat of fusion of ice: L_F = 80 cal/g.

Problem diagram:

Initially, the system is at a temperature of 0 °C, after receiving energy by heat the temperature remains the same, but the volume decreases, this indicates that there has been a phase transition (Figure 1). As the density of ice is less than the density of water (d_i < d_w) the volume occupied by the system initially is greater than the volume of the system after a mass m of ice has melted, the negative sign in the volume change indicates this situation.

Figure 1

Solution

a) The change in volume is given by the difference between the final volume of water, V_w, that formed and the initial volume of ice, V_i, that melted

\[ \begin{gather} \Delta V=V_{w}-V_{i} \tag{I} \end{gather} \]

The density of a substance is given by

\[ \bbox[#99CCFF,10px] {d=\frac{m}{V}} \]

writing this expression for ice and water

\[ \begin{gather} d_{i}=\frac{m}{V_{i}}\\ V_{i}=\frac{m}{d_{i}} \tag{II-a} \end{gather} \]

\[ \begin{gather} d_{w}=\frac{m}{V_{w}}\\ V_{w}=\frac{m}{d_{w}} \tag{II-b} \end{gather} \]

substituting expressions (II-a) and (II-b) into expression (I)

\[ \Delta V=\frac{m}{d_{w}}-\frac{m}{d_{i}} \]

factoring the mass m on the right-hand side of the equation and substituting the data of the problem

\[ \begin{gather} \Delta V=m\left(\frac{1}{d_{w}}-\frac{1}{d_{i}}\right)\\ -0.5=m\left(\frac{1}{1}-\frac{1}{0.92}\right) \end{gather} \]

multiplying the numerator and denominator of the first term in parentheses by 0.92

\[ \begin{gather} -0.5=m\left(\frac{1}{1}\times \frac{0.92}{0.92}-\frac{1}{0.92}\right)\\[5pt] -0.5=m\left(\frac{0.92}{0.92}-\frac{1}{0.92}\right)\\[5pt] -0.5=m\left(\frac{0.92-1}{0.92}\right)\\[5pt] -0.5=m\left(\frac{-{0.08}}{0.92}\right)\\[5pt] 0.5=\frac{0.08}{0.92}m\\m=\frac{0.5\times 0.92}{0.08}\\[5pt] m=\frac{0,46}{0.08} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {m=5.75\;\text{g}} \]

b) The amount of energy supplied by heat to melt the ice is calculated by the expression of the latent heat

\[ \bbox[#99CCFF,10px] {Q=mL} \]

\[ \begin{gather} Q=mL_{F}\\ Q=5.75\times 80 \end{gather} \]

\[ \bbox[#FFCCCC,10px] {Q=460\;\text{cal}} \]

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