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Phys 101
CHAPTER 13 – THERMAL ENERGY
Mechanical Work and Heat
It takes a lot of mechanical work to raise the temperature of something. This is good. If the
opposite were true, scuffing our feet on the floor as we walk could raise the temperature so high
as to cook our feet. In Chapter 6 we learned that the work done on an object is equal to the force
on it multiplied by the distance along which the force is applied. So the frictional force
associated with scuffing your feet acting through the lengths of the scuffs does mechanical work.
The first law of thermodynamics tells us that this work is just as effective in raising the
temperature of our feet as an equivalent amount of heat.
Example: Imagine that you are sitting in a bathtub with 20 liters of water and that you decide
that the 40 °C water is too cold. You decide to raise the temperature of the water by moving
your arms back and forth through the water. Assuming that there is no radiative or conductive
heat loss from the water, estimate how long would it take you to raise the temperature of the
water by 1 °C?
Let's assume that you move your arms back and forth once every second, each swing is
1/2-m long, and that the average force F exerted is 40 N in each direction. Therefore the
work done during each back and forth swing is
W = Fd = 40N × 1m = 40 J .
Because this amount of work is done every second, the power being generated by moving
your arms is P = E/∆t = 40 W.
In order to calculate the amount of heat it would take to raise the temperature of the
water, we need to know the mass m of the water. Since we know the volume of the water,
we can use the density of water to find m, ie
1000cm3
g
= 2 × 10 4 g
m = DV = 1 3 × 20 L ×
cm
1L
Because it takes 1 cal to raise the temperature of 1 g of water by 1°C, the heat flow Q that
would raise the temperature of this water by 1°C is 2×104 J.
Dividing the total amount of thermal energy required by the rate at which energy is being
generated gives us the required time,
E 2 × 104 cal 4.19 J
∆t = =
×
= 2090 s = 34.8 min
P
40J / s
1 cal
Notice that we had to convert calories to joules to get the units to cancel correctly.
Practice: How would the time change if you decided to push with twice the force? (Ans: It
would take half as long, 17.4 min.)
Specific Heat
The amount of heat (or other forms of work and energy) it takes to raise the temperature of
1 gram of any substance by 1°C is called the specific heat of the substance. So for water, the
specific heat is 1.0 cal/g°C since it takes 1 calorie to raise the temperature of 1 gram by 1°C. To
change the temperature of m grams by ∆T, the required heat is just
Phys 101
Q = mc∆T
where c is the specific heat. Note that if we measured the mass in kg, the temperature in kelvins,
and the energy in joules, the specific heat of water in another set of units would be 4186 J/kg⋅K.
Table 9-1 in the text gives values for the specific heats for some common materials.
Example: How much heat is required to raise the temperature of a 25-g aluminum cube from
60°C to 80°C?
Note that the temperature change is ∆T = 80°C – 60°C = 20°C. Looking up the value of the
specific heat of aluminum in the text, we find
Q = mc∆T = 25 g × 0.215 cal / g°C × 20°C = 108 cal
Practice: How much heat would it take to raise the temperature of a 50-g aluminum cube from
60°C to 80°C? (Ans: Since we have to twice the mass it should take twice the amount of energy,
ie 216 cal)
Change of State
The latent heat for a substance is the amount of energy required to change the physical state of a
unit of mass of that substance. For example, when water changes from a solid (ice) to a liquid,
80 calories per gram (334 kJ per kg) are required to break the bonds that hold the water
molecules tightly together. The values for the latent heat given in Table 9-2 in the text show that
the values for the latent heat vary over a large range of values. Note that in the reverse process,
the same amount of energy is given off. Thus this table also gives the latent heat of solidification
or condensation.
Example: A 20-kg block of ice at 0°C melts to form a puddle of water at 0°C. How much
energy does the ice absorb?
The latent heat of melting for water is 334 kJ/kg, so the heat energy needed to change the
physical state of the block from ice to liquid water is
kJ
Q = 20 kg × 334
= 6, 680 kJ
kg
Although the puddle is still at the same temperature, it has more energy. The energy
went into breaking the molecular bonds.
Practice: If there was half as much ice, how much energy would be absorbed? (Ans: 3340 kJ,
just half as much energy)
Example: If the process is reversed, that is the puddle at 0°C is frozen, how much energy is
given off by (ie, how much energy is removed from) the water.
The same amount of energy is given off as was absorbed in the previous example. This is
like a spring that has been stretched and at some later time, will be released. The energy
is stored in the spring and then released.