Note 26 Effects of Heat
When heat flows to and from an object, two things can happen.
temperature changed, and the object can have its phase changed.
The object can have its
Heat Capacity
How much heat is required to change the temperature of an object by 1 K is represented by “C”.
The object can be anything, for example, a car or a house.
C =
Q
ΔT
Alternatively, the amount of heat required to change the temperature of an object by the
temperature ∆T is
Q = C ΔT
The unit of heat capacity is J/K.
Specific Heat
How much heat is required to change the temperature of 1 kg of a substance by 1 K is
represented by “c”. The substance must be a pure substance, for example, iron or water.
c=
Q
mΔT
The amount of heat required to change a certain mass of a substance by the temperature ∆T is
Q = mcΔT
The unit of specific heat is J/(kg•K). Here are the specific heats for some substances.
Substance
c [J/(kg K)]
Water
4186
Ice
2090
Steam
2010
Air
1004
Iron
448
Copper
387
Lead
128
page 1
Calorimetry
Calorimetry is the process of measuring the flow of heat between objects, usually using a
calorimeter. Heat flows when there is a temperature difference. Heat flow stops when there is no
temperature difference (thermal equilibrium).
If we assume that no heat is lost to the surrounding due to an insulated calorimeter, then the
thermal energy is conserved. This means all of the heat flows add up to zero.
Q1 + Q2 + Q3 + ... = 0
In other words, the thermal energy lost by the hot objects plus the thermal energy gained by the
cold objects is zero. Remember that the heat flowing out is negative. Thus, all of the Qhot’s are
negative and all of the Qcold’s are positive..
Qhot,1 + Qhot,2 + Qhot,3 + ... = −(Qcold,1 + Qcold,2 + Qcold,3 + ...)
Example: Mixing Metal with Water 1
A 1.5 kg block of a certain type of metal heated to 100 °C is dropped into an insulated container
with 2.0 kg water at 20.0 °C. If the final temperature of the system is 35 °C, what is the specific
heat of the metal?
The heat flowing in to the hotter metal is
Qmetal = mcΔT = (1.5 kg)c(35 °C − 100 °C ) = (−97.5 kg °C )c
The heat flowing in to the colder water is
Qwater = mcΔT = (2.0 kg)(4186
J
kg⋅K
)(35 °C − 20 °C ) = 125580 J
Using the conservation of thermal energy,
(−97.5)c + 125580 = 0
⇒ c = 1288
J
kg⋅K
Example: Mixing Metal with Water 2
A 1.5 kg block of copper heated to 100 °C is dropped into an insulated container containing 2.0 kg
of water at 25 °C. What is the final temperature of the mixture?
The heat flowing in to the hotter metal is
Qmetal = mcΔT = (1.5 kg)(387
J
kg⋅K
)(Tf − 100 °C ) = 580.5(Tf − 100)
The heat flowing in to the colder water is
Qwater = mcΔT = (2.0 kg)(4186
J
kg⋅K
)(Tf − 25 °C ) = 8372(Tf − 25)
Using the conservation of thermal energy,
580.5(Tf − 100) + 8372(Tf − 25) = 0
8952.5Tf − 58050 − 209300 = 0
⇒ Tf = 29.9 °C
page 2
Phase Changes
Heat can also change the phase of a substance. Each phase of matter exists only under certain
set of conditions. We diagram out these conditions using the phase diagram. Here is an example
of a phase diagram. This is 1
Link
Because volume also plays a role, though mostly in the vapor phase (steam), this can be
visualized using three axes.
Link
page 3
Latent Heats
In phase transitions (moving across the lines separating the phases), thermal energy must be
added to or removed from the substance.
latent heat
of fusion, Lf
pressure
solid
latent heat
of vaporization, Lv
liquid
gas
temperature
latent heat
of sublimation, Ls
The amount of heat required to change phase are called latent heats. In general, the latent heat,
L, is the heat required to change the phase for every unit mass of a substance and it is measured
in J/kg.
L=
Q
m
Each phase change has its own latent heat. There are usually three of these. Melting and
freezing are described by the latent heat of fusion, Lf. Boiling and condensing are described by
the latent heat of vaporization, Lv. Certain substances can also sublimate, going from solid to gas
and vice versa. There is also a heat of sublimation.
Substance
Lf [J/kg]
Lv [J/kg]
Water
33.5 x 104
22.6 x 105
Copper
20.7 x 104
47.3 x 105
Ethyl Alcohol
10.8 x 104
8.55 x 105
Gold
6.28 x 104
17.2 x 105
Nitrogen
2.57 x 104
2.00 x 105
Lead
2.32 x 104
8.59 x 105
When going through a phase transition, the temperature does not change.
page 4
Here is a diagram of what happens to the temperature of water, including phase changes, as
thermal energy is added or removed.
temperature
condensing
100 °C
0 °C
–273 °C
boiling
freezing
thermal
energy
melting
ice
water
steam
page 5
Example: Steam burn
One gram of water and steam both at the temperature of 100 °C come into contact with skin at
35°C, how much heat is delivered to the skin assuming the final temperature of the water and
steam is also 35 °C?
For water, its temperature goes down to 35 °C. The heat removed from the water is
Qwater = mcΔT = (0.001 kg)(4186
J
kg⋅K
)(35 °C − 100 °C ) = −272 J
For steam, it has to be converted back to water first. The heat removed from the steam for this
process is
Q = mL = (0.001 kg)(−22.6 ×105
J
kg
) = −2260 J
Now that the steam has become water, the water cools to 35 °C. This requires a removal of 272 J
so the total heat removed from the steam is 2532 J which goes to the skin.
Example: Boiling water
You add 5.60x105 J of thermal energy to 0.220 kg of water that has an initial temperature of 50.0
°C. What is the final temperature of the steam?
The first thing the thermal energy tries to do is to bring the water to 100°C. The heat required for
this is
Qto boil = mcΔT = (0.220)(4186)(50) = 4.60 ×104 J
There is more than enough thermal energy to do this. After completing the above process, the
remaining thermal energy is
Qremain,1 = Qinitial −Qto boil = 5.60 ×105 − 4.60 ×104 = 5.14 ×105 J
Once the water is at 100 °C, the next thing the thermal energy does is to generate steam. The
heat required to turn all of the water into steam is
Qto evaporate = mLv = (0.220)(22.6 ×105 ) = 4.97 ×105 J
There is still some thermal energy remaining after this process.
Qremain,2 = Qremain,1 −Qto evaporate = 5.14 ×105 − 1.94 ×105 = 1.68 ×104 J
This goes to heating up the steam.
Qto heat = 5.14 ×105 J = mcΔT = (0.220)(2010)ΔT
⇒ ΔT = 37.9 K
The final temperature of the steams is
ΔT = Tf −Ti
⇒ Tf = 138 °C
page 6
Example: Mixing ice and water
An 10 g ice cube at –10 °C is placed in an insulated cup with 100 g of water at 25 °C. What is the
final state of the mixture?
The same concept applies to this problem as before, namely,
Qice + Qwater = 0
However, due to phase transitions, each of these two terms can contain multiple terms.
Therefore, we may have to break this up into parts.
The questions to ask are how much heat is required to do the next possible process. For the ice,
the first question is how much heat is required to heat it to 0 °C? Next question would be how
much heat is required to melt the ice? The final question is how much heat is required to heat the
melted ice to its final temperature?
temperature
25 °C
ice
thermal
energy
0 °C
water
–10 °C
Moreover, you have to, at the same time, ask what could be happening to the water. Here are the
possible processes for the water.
temperature
25 °C
ice
thermal
energy
0 °C
water
–10 °C
Somewhere between these two possibilities is thermal equilibrium. Let’s start.
1. First set of possible processes. Ice warms to 0 °C and water cools to 0 °C.
Qice to 0 = mcΔT = (0.01 kg)(2050
Qwater to 0 = mcΔT = (0.1 kg)(4186
J
kg⋅K
)(10 K ) = 205 J
J
kg⋅K
)(−25 K ) = −10, 465 J
The result here is that all of the ice will warm to 0 °C while water temperature drop a little. Let’s
see what the water temperature is. The water will lose an amount of thermal energy equal to what
is necessary to warm the ice. This is –205 J.
−205 = (0.1 kg)(4186
J
kg⋅K
)(Tf − 25 °C ) ⇒ Tf = 24.51 °C
page 7
2. Second set of possible processes. Ice melts and water cools to 0 *C.
Qice melts = mL = (0.01 kg)(33.5 ×104
J
kg
) = 3, 350 J
Qwater to 0 = −10, 260 J
Comparing the heats, the water has enough thermal energy to melt the ice. The ice will gain a
total of 3,355 J of heat and the water will lose 3,355 J of heat. The final temperature of the water
is
−3, 355 J = mcΔT = (0.1 kg)(4,186
J
kg⋅K
)(Tf − 24.51 °C ) ⇒ Tf = 16.51 °C
3. Once the two objects are in the same phase, they will end up with the same final temperature.
Qice + Qwater = 0
Notice the specific heat doesn’t matter here now that they are both water. The final temperature
of the mixture is
(0.01 kg)(4186
J
kg⋅K
)(Tf − 0 °C ) + (0.1 kg)(4186
0.01Tf + 0.1Tf − (0.1)(16.5 °C ) = 0
J
kg⋅K
)(Tf − 16.51 °C ) = 0
⇒ Tf = 15.0 °C
Here is all of the processes summarized in diagram form.
temperature
15 °C
ice
thermal
energy
water
The reason for the small change in the water’s temperature is because its mass is much larger
than that of the ice and water has a larger specific heat.
page 8
Example: Hot iron
A 50 g block of iron heated to 800 °C is placed into a closed, and insulated container containing
100 g of water at 80 °C. What are the final phases of the water and its temperature?
This is what we start with: iron 50 g @ 800 °C and water 100 g @ 80 °C.
1. The first possible process here is the heating of the water to 100 °C and the iron dropping to
100 °C which will not happen. This requires an amount of thermal energy equal to
Qwater to 100 = mcΔT = (0.1 kg)(4186
Qiron to 100 = mcΔT = (0.05 kg)(448
J
kg⋅K
J
kg⋅K
)(100 °C − 80 °C ) = 8, 372 J
)(100 °C − 800 °C ) = −15, 680 J
So the water will all go up to 100 °C. The iron will lose thermal energy and drop to
−8, 372 J = (0.05 kg)(448
J
kg K
)(Tf − 800 °C ) ⇒ Tf = 426.25 °C
Since the temperatures are still different, we keep going.
2. The second possible process is if the water all evaporate and the iron drops to 100 °C.
Q = mLv = (0.1 kg)(22.6 ×105
J
kg
) = 2.26 ×105 J
Qiron to 100 = mcΔT = (0.05 kg)(448
J
kg⋅K
)(100 °C − 426.25 °C ) = −7, 308 J
The iron becomes the limit here. The lowest temperature that the iron will drop to is 100 °C.
Therefore, the only question is how much water turns into steam. With this much heat going into
the water, the amount of steam generated is
Q = mLv = m(22.6 ×105
J
kg
) = +7, 302 J
m = 3.2 ×10−3 kg = 3.2 g
Here is the final state. The iron is 50 g @ 100 °C. The water remaining is 96.8 g @ 100 °C. The
rest of the water went into steam, 3.2 g @ 100 °C. Notice all of the temperatures are the same as
required by thermal equilibrium.
page 9