C/P 02 · Heat, phase change and entropy
Warming Metal and Melting Ice
Use calorimetry and a heating curve to find specific heat and latent heat.
Passage
Heat is energy transferred between objects at different temperatures. When a substance absorbs heat q without changing phase, its temperature change ΔT is given by q = mcΔT, where m is the mass and c is the specific heat of the substance. During a phase change at constant pressure, the substance absorbs or releases heat at constant temperature, and the heat is related to the mass by q = mL, where L is the latent heat of the transition. Both relationships are used in calorimetry, in which the heat lost by a hot object is assumed to equal the heat gained by a cold one.
Calorimetry has practical limits. No cup is perfectly insulated, the thermometer and the cup absorb a little heat of their own, and the final temperature has to be read after the contents have mixed. A careful experimenter checks whether the measured result is consistent with an isolated system or whether some heat was probably exchanged with the room.
Water has an unusually large specific heat among common substances, which is one reason that large bodies of water and living tissue change temperature slowly. Its latent heat of vaporization is also large, which is why the evaporation of sweat is an effective way to cool the skin. In the laboratory, the same ideas are used in two ways: mixing a heated sample into water to find its specific heat, and recording a heating curve for a pure substance, in which flat regions mark phase changes and the slope of each sloped region depends on the specific heat of that phase.
Part 1. A student heats a 100 g metal sample to 80 °C and drops it into 100 g of water at 20 °C in an insulated cup. After the contents are stirred, the temperature stops changing. The specific heat of water is 4 J/(g·K). Table 1 shows the results.
Table 1. Calorimetry results for the metal and the water.
| Object | Mass (g) | Initial temperature (°C) | Final temperature (°C) |
|---|---|---|---|
| Metal | 100 | 80 | 30 |
| Water | 100 | 20 | 30 |
Part 2. In a second experiment, a 50 g block of ice at −20 °C is heated by an electric heater that delivers energy at a constant rate of 100 J/s. The heater is in direct contact with the sample, the pressure stays constant, and the temperature is recorded as the sample is heated. Table 2 shows the results.
The sample sits in a thin-walled sealed container so that no material is lost, and a thermometer in contact with the sample is read at the times listed. The student assumes that all of the heater’s output goes into the sample and that the container itself absorbs a negligible amount of heat.
Table 2. Temperature of the 50 g sample during heating.
| Time (s) | Temperature (°C) | Observation |
|---|---|---|
| 0 | −20 | Solid ice |
| 20 | 0 | Ice begins to melt |
| 100 | 0 | Ice and liquid water |
| 180 | 0 | Last ice melts |
| 220 | 20 | Liquid water only |
The student also notes that the melting happened at a single, steady temperature, and that the sample warmed faster as a solid than it did as a liquid. For the entropy question below, take the melting temperature to be about 270 K.
Questions
Question 1 of 6
Assuming the cup in Part 1 is perfectly insulated, what is the specific heat of the metal?
This would mean the metal gave up only 2000 J, half of what the water gained. In an isolated system the two amounts of heat are equal in size.
This would mean the metal gave up 8000 J, twice what the water gained. The heat lost by the metal equals the heat gained by the water.
This is the specific heat of water. The metal changed temperature by 50 K and the water by only 10 K, with equal masses, so the metal’s specific heat is one fifth of water’s.
Why the answer is what it is
The water gains q = mcΔT = 100 × 4 × 10 = 4000 J. The metal loses the same 4000 J while cooling by 50 K: 4000 = 100 × c × 50, so c = 0.8 J/(g·K).
Review: Work, Heat, and the First Law of Thermodynamics · Thermodynamics
Question 2 of 6
Suppose some heat escaped from the cup to the room during Part 1, but the student still assumed that the heat gained by the water equals the heat lost by the metal. The calculated specific heat of the metal would be:
Heat loss lowers the final temperature, which makes the water’s rise smaller and the metal’s drop larger. Both changes push the calculated value down, not up.
The direction of the error can be reasoned out without that number. A smaller measured rise in the water gives a smaller calculated heat transfer.
The heat lost to the room does not cancel. The assumption of equal heats is then wrong, and the calculation inherits the error.
Why the answer is what it is
With heat loss, the final temperature is lower than it would be in an isolated system. The water’s measured rise (the quantity that sets the calculated heat) is smaller, and the metal’s measured drop is larger. Dividing a smaller heat by a larger ΔT gives a calculated specific heat that is too low.
Review: Work, Heat, and the First Law of Thermodynamics · Thermodynamics
Question 3 of 6
In Table 2 the temperature stays at 0 °C from 20 s to 180 s even though the heater keeps delivering energy. Which statement best explains this?
The heater runs at a constant rate throughout the experiment, as stated in Part 2.
If the specific heats were the same the temperature would still rise. A flat plateau means the energy is going into something other than temperature.
This would have to hold only during the phase change and not before or after it. The sample warms steadily on both sides of the plateau.
Why the answer is what it is
Temperature reflects the average kinetic energy of the molecules. During melting, the energy goes into breaking the ordered arrangement of the solid, which raises potential energy and leaves the temperature unchanged until all of the solid has melted.
Review: States of Matter & Phase Transitions · Work, Heat, and the First Law of Thermodynamics
Question 4 of 6
What is the latent heat of fusion of this substance, based on Table 2?
This would come from a plateau of only 80 s. The plateau lasts from 20 s to 180 s, which is 160 s.
This would need 32,000 J, which is twice the energy delivered during the plateau.
16,000 J is the total energy absorbed during melting. Latent heat is the energy per gram, so divide by the 50 g mass.
Why the answer is what it is
The plateau lasts 160 s at 100 J/s, so the sample absorbs 16,000 J. Using q = mL, L = 16,000 J ÷ 50 g = 320 J/g.
Review: States of Matter & Phase Transitions · Work, Heat, and the First Law of Thermodynamics
Question 5 of 6
From the data in Table 2, what is the ratio of the specific heat of the liquid to the specific heat of the solid?
This is inverted. A substance with a larger specific heat warms more slowly for the same energy input, and the liquid warmed more slowly.
The two phases warm at different rates (1 K/s as solid, 0.5 K/s as liquid), so their specific heats are not the same.
The warming rates differ by a factor of 2, not 4. Specific heat is inversely proportional to the rate, so the ratio is also 2.
Why the answer is what it is
The same power heats the same mass in both phases, so the specific heat is inversely proportional to the warming rate. The solid warms 20 K in 20 s (1 K/s) and the liquid 20 K in 40 s (0.5 K/s). The liquid warms at half the rate, so its specific heat is twice as large.
Review: Work, Heat, and the First Law of Thermodynamics · Thermodynamics
Question 6 of 6
Which statement best describes the entropy change of the sample while it melts at about 270 K?
Absorbing heat reversibly gives a positive q, and ΔS = q/T has the same sign. The liquid is also more disordered than the solid.
Entropy change depends on heat divided by temperature, not on the temperature change. A phase change at constant T still has ΔS = q/T.
Entropy is heat divided by temperature. Dividing 16,000 J by about 270 K gives roughly 60 J/K.
Why the answer is what it is
For a reversible phase change at constant temperature, ΔS = q/T = 16,000 J ÷ 270 K, which is about 60 J/K. The sign is positive because the liquid is more disordered than the solid.