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Physics LibreTexts

13.6: Examples

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Example 13.6.1: Calorimetry

The specific heat of aluminum is 900 J/kg·K, and that of water is 4186 J·K. Suppose you drop a block of aluminum of mass 1 kg at a temperature of 80C in a liter of water (which also has a mass of 1 kg) at a temperature of 20C. What is the final temperature of the system, assuming no exchange of heat with the environment takes place? How much energy does the aluminum lose/the water gain?

Solution

Let us call TAl the initial temperature of the aluminum, Twater the initial temperature of the water, and Tf their final common temperature. The thermal energy given off by the aluminum equals ΔEAl=CAl(TfTAl) (this follows from the definition (13.2.1) of heat capacity; we could equally well call this quantity “the heat given off by the aluminum”). In the same way, the thermal energy change of the water (heat absorbed by the water) equals ΔEwater=Cwater(TfTwater). If the total system is closed, the sum of these two quantities, each with its appropriate sign, must be zero:

0=ΔEAl+ΔEwater=CAl(TfTAl)+Cwater(TfTwater).

This equation for Tf has the solution

Tf=CAlTAl+CwaterTwaterCAl+Cwater.

As you can see, the result is a weighted average of the two starting temperatures, with the corresponding heat capacities as the weighting factors.

The heat capacities C are equal to the given specific heats multiplied by the respective masses. In this case, the mass of aluminum and the mass of the water are the same, so they will cancel in the final result. Also, we can use the temperatures in degrees Celsius, instead of Kelvin. This is not immediately obvious from the final expression (???), but if you look at (???) you’ll see it involves only temperature differences, and those have the same value in the Kelvin and Celsius scales.

Substituting the given values in (???), then, we get

Tf=900×80+4186×20900+4186=30.6C.

This is much closer to the initial temperature of the water, as expected, since it has the greater heat capacity. The amount of heat exchanged is

Cwater (TfTwater )=4186×(30.620)=44,440J=44.4kJ.

So, 1 kg of aluminum gives off 44.4 kJ of thermal energy and its temperature drops almost 50C, from 80C to 30.6C, whereas 1 kg of water takes in the same amount of thermal energy and its temperature only rises about 10.6C.

Example 13.6.2: Equipartition of energy

Estimate the speed of an oxygen molecule in air at room temperature (about 300 K).

Solution

Recall that in Section 13.2 I mentioned that the average translational kinetic energy of a molecule in a system at a temperature T is 32kBT (Equation (13.2.7), where kB, Boltzmann’s constant, is equal to 1.38 × 10−23 J/K. So, at T = 300 K, a molecule of oxygen (or of anything else, for that matter) should have, on average, a kinetic energy of

Ktrans=32kBT=32×1.38×1023×300J=6.21×1021J.

Since K=12mv2, we can figure out the average value of v2 if we know the mass of an oxygen molecule. This is something you can look up, or derive like this: One mole of oxygen atoms has a mass of 16 grams (16 is the atomic mass number of oxygen) and contains Avogadro’s number of atoms, 6.02 × 1023. So a single atom has a mass of 0.016 kg/6.02 × 1023 = 2.66 × 10−26 kg. A molecule of oxygen contains two atoms, so it has twice the mass, m = 5.32 × 10−26 kg. Then,

v2=2Ktransm=2×6.21×1021J5.32×1026kg=2.33×105m2s2.

The square root of this will give us what is called the “root mean square” velocity, or vrms:

vrms=2.33×105m2s2=483ms.

This is of the same order of magnitude as (but larger than) the speed of sound in air at room temperature (about 340 m/s, as you may recall from Chapter 12).


This page titled 13.6: Examples is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Julio Gea-Banacloche (University of Arkansas Libraries) via source content that was edited to the style and standards of the LibreTexts platform.

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