Q4Q Question: A sample聽A聽of liquid... [FREE SOLUTION]

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Chapter 18: Q4Q (page 540)

Question: A sampleAof liquid water and a sampleBof ice, of identical mass, are placed in a thermally insulated container and allowed to come to thermal equilibrium. Figure 18-25ais a sketch of the temperatureTof the samples versus timet. (a) Is the equilibrium temperature above, below, or at the freezing point of water? (b) In reaching equilibrium, does the liquid partly freeze, fully freeze, or undergo no freezing? (c) Does the ice partly melt, fully melt, or undergo no melting?

Short Answer

Expert verified

a)The equilibrium temperature isat freezing point.
b)In reaching equilibrium, the liquidwater undergoes no freezing.
c) The ice partly melts.

Step by step solution

The given data

i)The mass of sample A of liquid water and the mass of sample B of ice are identical.
ii) The graph of temperature versus time.

Understanding the concept of thermal equilibrium

When two substances in physical contact with one another are in thermal equilibrium, no heat energy is exchanged. Observing the graph (a) of the figure 18-25, we can find the equilibrium temperature is above, below, or at the freezing point of the water. Also, we can find whether in reaching equilibrium, the liquid partly freezes, fully freezes, or undergoes no freezing and whether the ice partly melts, fully melts, or undergoes no melting.

Formula:

The amount of energy transferred by the body, $Q = m c 鈭� T$ 鈥�(颈)

Where,
m is mass, c is specific heat and $鈭� T$ is change in temperature

(a) Calculation of the point of the equilibrium temperature

We note that there is a rise in temperature of sample B of ice that equals to the loss in the temperature of sample A of liquid water and the temperature of sample A and B equals at equilibrium. Thus, at equilibrium point ${(T_{f} - T_{l})}_{A} = {(T_{f} - T_{l})}_{s}$

Where $T_{f}$ is the final temperature, $T_{l}$ is initial temperature and ${(T_{f} - T_{l})}_{A} a n d {(T_{f} - T_{l})}_{s}$ of sample A and B respectively.

Let $Q_{A} a n d Q_{B}$ be the energies transformed in sample A and B respectively and $C_{w a t e r} a n d C_{I c e}$ and be the specific heats of the water and ice respectively. Then,

The energy transformed $Q_{A}$ is given using equation (i) by

role="math" localid="1663066457593" $Q_{A} = C_{w a t e r} m {(T_{f} - T_{1})}_{A}$

And theenergy transformedis given using equation (i) by

$Q_{B} = C_{i c e} m {(T_{f} - T_{1})}_{B}$

But, specific heat of water is $C_{w a t e r} = 4.187 K J / k g K$

And specific heat of ice is $C_{I c e} = 2.108 K J / k g K$

Thus for all other values being constant, the value of heat is $Q_{A} > Q_{B}$ .

From the Graph (a) of Figure 18-25, we can observe that at the equilibrium, the horizontal line is at the same temperature, that is, at the point where the two-graph meet.

Hence, the equilibrium temperature is at freezing point.

(b) Calculation of the condition of water after reaching the equilibrium point

From Graph (a) of Figure 18-25, we can observe that there is no horizontal section for the water graph before it meets with the ice graph.

Hence, water undergoes no freezing.

(c) Calculation of the condition of ice at the equilibrium temperature

From Graph (a) of Figure 18-25, we can observe the horizontal line for the ice graph before meeting point, but it is interrupted by the thermal equilibrium, and hence, the phase change is not completed.

Therefore, we can say that the ice partly melts.