Question

When ice with a temperature below $0 ° C$ is mixed with water with a temperature above $0 ° C$ , there are four possibilities: the ice melts and the final temperature is above $0 ° C$ , the water freezes and the final temperature is below $0 ° C$ , part of the ice melts and the temperature of the mixture becomes $0 ° C$ , and part of the water freezes and the temperature of the mixture becomes $0 ° C$ . On the horizontal axis we lay off the amount of heat that the water gives off in cooling and freezing (the upper straight lines) and the amount of heat that the ice absorbs in heating and melting (the lower straight lines). The scale along the horizontal axis is arbitrary, that is, the scale value is not specified. The temperature (in degrees Celsius) is laid off on the vertical axis. Find the final result of mixing whose beginning is shown in each figure. When either all the water freezes or all the ice melts, determine the final temperature.

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Accepted Answer

Each figure accompanying the problem contains the initial segments of the graphs representing the cooling of water or the heating of ice. Continuing these graphs, we arrive at the intersection point in each figure.

If the point of intersection lies above the horizontal line corresponding to a temperature of $0 ° C$ , the final temperature is positive, when the point lies below this line, the final temperature is negative. If the graphs meet on the line $t = 0 ° C$ , the final temperature is $0 {}^{\circ}C$ and the amount of the phase that has a horizontal section on the graph prior to intersection will decrease. The ratio of the length of this section to the total length of the horizontal section corresponding to this phase determines the fraction of the initial mass of this phase that has transformed into the other phase. When analysing the graphs, we must bear in mind that the slopes of the straight lines are determined by the mass of water or ice and their specific heat capacity by the formula $\frac{Δ t}{Δ Q} = \frac{1}{c m} .$
Here one must bear in mind that the specific heat capacity of water is twice as high as that of ice. The length of the horizontal sections corresponding to the water freezing or the ice melting is determined by the fact that the amount of heat required for melting a certain amount of ice is equal to the amount of heat required for heating the same mass of water to $80 ° C$ .

(a)

For the sake of illustration, Figure (a) accompanying the answer shows the diagram for the cooling off of a mass of water from $80$ to $0 ° C$ , then the freezing of this water, and finally the cooling off of the ice down to $t = - 40 ° C$ .

Figure (b) accompanying the answer shows the reverse process in which the same amount of ice is heated from $- 80$ to $0 {}^{\circ}C$ , then melted, and finally heated in the form of water to $60 ° C$ . The scales along the horizontal axes are arbitrary but equal, with the amount of heat expressed in arbitrary units. (It is easy to see that all this has no effect on the answer.)

The two diagrams are combined in Figure (c) accompanying the answer. In the present case we see that the final temperature is $0 ° C$ and half of the ice has melted. Applying this procedure to the case illustrated by Figure (a) accompanying the problem, we see that the ice has completely melted and the final temperature is $10 ° C$ ; for Figure (b) accompanying the problem, half of the ice has melted and the final temperature is $0 {}^{\circ}C$ ; for Figure (c) accompanying the problem, the case is similar to (b) but half of the water has frozen; finally, for Figure (d), all the water has frozen and the final temperature is $- 20 ° C$ .

L.A.Sena Solutions for Chapter: Molecular Physics and Thermodynamics, Exercise 1: Exercise

L.A.Sena Physics Solutions for Exercise - L.A.Sena Solutions for Chapter: Molecular Physics and Thermodynamics, Exercise 1: Exercise

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