19.6 Calorimetry

At one time or another you’ve probably put an ice cube into a hot drink to cool it quickly. You were engaged, in a somewhat trial-and-error way, in a practical aspect of heat transfer known as calorimetry.

FIGURE 19.15 shows two systems thermally interacting with each other but isolated from everything else. Suppose they start at different temperatures $T_{1}$ and $T_{2}$ . As you know from experience, heat energy will be transferred from the hotter to the colder system until they reach a common final temperature $T_{f}$ . The systems will then be in thermal equilibrium and the temperature will not change further.

The figure illustrates the thermal interaction between two systems.

FIGURE 19.15 Full Alternative Text

The insulation prevents any heat energy from being transferred to or from the environment, so energy conservation tells us that any energy leaving the hotter system must enter the colder system. That is, the systems exchange energy with no net loss or gain. The concept is straightforward, but to state the idea mathematically we need to be careful with signs.

Let $Q_{1}$ be the energy transferred to system 1 as heat. Similarly, $Q_{2}$ is the energy transferred to system 2. The fact that the systems are merely exchanging energy can be written $| Q_{1} | = | Q_{2} |$ . The energy lost by the hotter system is the energy gained by the colder system, so $Q_{1}$ and $Q_{2}$ have opposite signs: $Q_{1} = - Q_{2}$ . No energy is exchanged with the environment, hence it makes more sense to write this relationship as

Q_{net} = Q_{1} + Q_{2} = 0

(19.21)

This idea is not limited to the interaction of only two systems. If three or more systems are combined in isolation from the rest of their environment, each at a different initial temperature, they will all come to a common final temperature that can be found from the relationship

Q_{net} = Q_{1} + Q_{2} + Q_{3} + \dots = 0

(19.22)

NOTE The signs are very important in calorimetry problems. $Δ T$ is always $T_{f} - T_{i}$ , so $Δ T$ and $Q$ are negative for any system whose temperature decreases. The proper sign of $Q$ for any phase change must be supplied by you, depending on the direction of the phase change.

PROBLEM-SOLVING STRATEGY 19.2
Calorimetry problems

MODEL Model the systems as interacting with each other but isolated from the larger environment.

VISUALIZE List known information and identify what you need to find. Convert all quantities to SI units.

SOLVE The mathematical representation, a statement of energy conservation, is

Q_{net} = Q_{1} + Q_{2} + \dots = 0

For systems that undergo a temperature change, $Q = M c (T_{f} - T_{i})$ . Be sure to have the temperatures $T_{i}$ and $T_{f}$ in the correct order.
For systems that undergo a phase change, $Q = \pm M L$ . Supply the correct sign by observing whether energy enters or leaves the system.
Some systems may undergo a temperature change and a phase change. Treat the changes separately. The heat energy is $Q = Q_{Δ T} + Q_{phase}$ .

REVIEW Is the final temperature in the middle? $T_{f}$ that is higher or lower than all initial temperatures is an indication that something is wrong, usually a sign error.

Workbook Exercise 15

NOTE You may have learned to solve calorimetry problems in other courses by writing $Q_{gained} = Q_{lost}$ . That is, by balancing heat gained with heat lost. That approach works in simple problems, but it has two drawbacks. First, you often have to “fudge” the signs to make them work. Second, and more serious, you can’t extend this approach to a problem with three or more interacting systems. Using $Q_{net} = 0$ is much preferred.

EXAMPLE 19.5 Calorimetry with a phase change

Your 500 mL soda is at $20 ° C,$ room temperature, so you add 100 g of ice from the $- 20 ° C$ freezer. Does all the ice melt? If so, what is the final temperature? If not, what fraction of the ice melts? Assume that you have a well-insulated cup.

MODEL We have a thermal interaction between the soda, which is essentially water, and the ice. We need to distinguish between three possible outcomes. If all the ice melts, then $T_{f} > 0 ° C$ . It’s also possible that the soda will cool to $0 ° C$ before all the ice has melted, leaving the ice and liquid in equilibrium at $0 ° C$ . A third possibility is that the soda will freeze solid before the ice warms up to $0 ° C$ . That seems unlikely here, but there are situations, such as the pouring of molten metal out of furnaces, when all the liquid does solidify. We need to distinguish between these before knowing how to proceed.

VISUALIZE All the initial temperatures, masses, and specific heats are known. The final temperature of the combined $soda + ice$ system is unknown.

SOLVE Let’s first calculate the heat needed to melt all the ice and leave it as liquid water at $0 ° C$ . To do so, we must warm the ice to $0 ° C,$ then change it to water. The heat input for this two-stage process is

Q_{melt} = M_{i} c_{i} (20 K) + M_{i} L_{f} = 37, 500 J

where $L_{f}$ is the heat of fusion of water. It is used as a positive quantity because we must add heat to melt the ice. Next, let’s calculate how much heat energy will leave the soda if it cools all the way to $0 ° C$ . The volume is $V = 500 mL = 5.00 \times 10^{- 4} m^{3}$ and thus the mass is $M_{s} = ρ V = 0.500 kg$ . The heat loss is

Q_{cool} = M_{s} c_{w} (- 20 K) = - 41, 900 J

where $Δ T = - 20 K$ because the temperature decreases. Because $| Q_{cool} | > Q_{melt}$ , the soda has sufficient energy to melt all the ice. Hence the final state will be all liquid at $T_{f} > 0$ . (Had we found $| Q_{cool} | < Q_{melt}$ , then the final state would have been an ice-liquid mixture at $0 ° C$ .)

Energy conservation requires $Q_{ice} + Q_{soda} = 0$ . The heat $Q_{ice}$ consists of three terms: warming the ice to $0 ° C,$ melting the ice to water at $0 ° C,$ then warming the $0 ° C$ water to $T_{f}$ . The mass will still be $M_{i}$ in the last of these steps because it is the “ice system,” but we need to use the specific heat of liquid water. Thus

Q_{ice} + Q_{soda} = [M_{i} c_{i} (20 K) + M_{i} L_{f} + M_{i} c_{w} (T_{f} - 0 ° C)] + M_{s} c_{w} (T_{f} - 20 ° C) = 0

We’ve already done part of the calculation, allowing us to write

37, 500 J + M_{i} c_{w} (T_{f} - 0 ° C) + M_{s} c_{w} (T_{f} - 20 ° C) = 0

Solving for $T_{f}$ gives

T_{f} = \frac{20 M_{s} c_{w} - 37, 500 J}{M_{i} c_{w} + M_{s} c_{w}} = 1.7 ° C

REVIEW As expected, the soda has been cooled to nearly the freezing point.

EXAMPLE 19.6 Three interacting systems

A 200 g piece of iron at $120 ° C$ and a $150 g$ piece of copper at $- 50 ° C$ are dropped into an insulated beaker containing 300 g of ethyl alcohol at $20 ° C$ . What is the final temperature?

MODEL Here you can’t use a simple $Q_{gained} = Q_{lost}$ approach because you don’t know whether the alcohol is going to warm up or cool down. In principle, the alcohol could freeze or boil, but the masses and temperatures of the metals suggest that the temperature will not change greatly. We will assume that the alcohol remains a liquid.

VISUALIZE All the initial temperatures, masses, and specific heats are known. We need to find the final temperature.

SOLVE Energy conservation requires

Q_{i} + Q_{c} + Q_{e} = M_{i} c_{i} (T_{f} - 120 ° C) + M_{c} c_{c} (T_{f} - (- 50 ° C)) + M_{e} c_{e} (T_{f} - 20 ° C) = 0

Solving for $T_{f}$ gives

T_{f} = \frac{120 M_{i} c_{i} - 50 M_{c} c_{c} + 20 M_{e} c_{e}}{M_{i} c_{i} + M_{c} c_{c} + M_{e} c_{e}} = 26 ° C

REVIEW The temperature is between the initial iron and copper temperatures, as expected, and well below the boiling temperature of the alcohol, validating our assumption of no phase change. It turns out that the alcohol warms up $(Q_{e} > 0)$ , but we had no way to know this without doing the calculation.