Example 14-6 Cooling Coffee
Your expensive coffee maker produces coffee at a temperature of 95.0°C (203°F), which you find is a bit too warm to
drink comfortably. To cool the coffee, you pour 0.350 kg of brewed coffee (mostly water) at 95.0°C into a 0.250-kg
aluminum cup that is initially at room temperature (20.0°C). What is the final temperature of the coffee and cup? ­Assume
that the coffee and cup are thermally isolated.
Set Up
There are three unknown quantities in this
problem: the quantity of heat Qcup that flows
into the cup; the quantity of heat Qcoffee that
flows into the coffee (which will be negative
since heat flows out of the coffee); and the
final temperature Tf of the two objects
(which is what we want to find). So we need
three equations that relate these quantities.
We can get two equations by writing
Equation 14-20 twice, once for the cup and
once for the coffee. To get a third equation,
we’ll use the idea that the two objects are
thermally isolated, so the energy that flows
out of the hot coffee must equal the energy
that flows into the cool aluminum cup. So
Qcoffee (for the coffee) and Qcup (for the cup)
both involve the same number of joules.
However, Q is negative for the coffee (heat
flows out of it) and positive for the cup (heat
flows into it), so Qcoffee is equal to the negative of Qcup.
Solve
We are given the masses and initial temperatures of both the cup and the coffee, and we
can get the specific heats from Table 14-3.
Write equations for the three unknowns
Qcup, Qcoffee, and Tf. Note that for each
object, T is the difference between the final
and initial temperatures for that object.
Quantity of heat and the
resulting temperature
change:
Q = mc T
Energy is conserved:
(14-20)
coffee
95.0°C
0.350 kg
cup
20.0°C
0.250 kg
coffee in cup;
temperature
Tf = ?
Qcoffee = 2Qcup
The cup has mass mcup = 0.250 kg and initial temperature Tcup,i =
20.0°C, is made of aluminum with cAl = 910 J # kg21 # K21, and ends up
at temperature Tf. Equation 14-20 for the cup says
Qcup = mcupcAl Tcup = mcupcAl(Tf 2 Tcup,i)
Note that the final temperature Tf will be greater than Tcup,i (the cup
gets warmer), so Tf 2 Tcup,i 7 0 and Qcup will be positive.
The coffee has mass mcoffee = 0.350 kg and initial temperature
Tcoffee,i = 95.0°C, is made almost completely of water with cwater =
4186 J # kg21 # K21, and ends up at the same final temperature Tf as the
cup. Equation 14-20 for the coffee says
Qcoffee = mcoffee cwater Tcoffee = mcoffee cwater(Tf 2 Tcoffee,i)
The final temperature Tfinal will be less than Tcoffee,i (the coffee gets
cooler), so Tf 2 Tcoffee,i 6 0 and Qcoffee will be negative.
The equation of energy conservation is
Qcoffee = 2Qcup
Combine these three equations to get a single
equation for Tf, the quantity we are trying to
find.
Substitute Qcup and Qcoffee from the first two equations into the third
equation for energy conservation:
Qcoffee = 2Qcup
mcoffee cwater(Tf 2 Tcoffee,i) = –mcupcAl(Tf 2 Tcup,i)
The only unknown quantity in this equation is the final
temperature Tf.
Solve for the final temperature Tf.
Multiply out both sides of the equation:
mcoffee cwaterTf 2 mcoffee cwaterTcoffee,i = 2mcupcAlTf + mcupcAlTcup,i
Rearrange so that all the terms with Tf are on the same side of
the equation:
mcoffee cwaterTf + mcup cAlTf = mcoffee cwaterTcoffee,i + mcup cAlTcup,i
or
(mcoffee cwater + mcupcAl) Tf = mcoffee cwaterTcoffee,i + mcup cAlTcup,i
Solve for Tf:
Tf =
=
mcoffeec waterTcoffee,i + mcup c AlTcup,i
mcoffee c water + mcup c Al
c
10.350 kg2 14186 J  kg -1 K -1 2 195.0C2
d
+ 10.250 kg2 1910 J kg -1 K -1 2 120.0C2
10.350 kg2 14186 J  kg -1  K -1 2 + 10.250 kg2 1910 J  kg -1  K -1 2
= 84.9°C
(Recall that a temperature change of 1°C is equivalent to a temperature
change of 1 K. This explains why we can exchange kelvins and degrees
Celsius in the above equation.)
The coffee cools from 95.0°C to 84.9°C (close to the ideal drinking
temperature preferred by coffee aficionados) and the aluminum cup
warms from 20.0°C to 84.9°C.
Reflect
The temperature of the coffee decreases and
the temperature of the cup increases, just as
we expected. We can check our results by
calculating Qcup and Qcoffee. This calculation
shows that Qcoffee = 2Qcup, which must be
true for energy to be conserved.
Note that the final temperature of
84.9°C is closer to the initial temperature
of the coffee (95.0°C) than to the initial
temperature of the cup (20.0°C). That’s
­because the mass and the specific heat are
both greater for the coffee than for the
cup, so a given quantity of heat produces
a smaller temperature change in the coffee
than in the cup.
The temperature change for the cup is
Tcup = Tf 2 Tcup,i = 84.9°C 2 20.0°C = +64.9°C = +64.9 K
(Recall that 1°C and 1 K represent the same temperature change.) The
heat that flows into the cup is
Qcup = mcupcAl(Tf 2 Tcup,i)
= (0.250 kg)(910 J # kg21 # K21)(+64.9 K)
= 1.48 * 104 J
The temperature change for the coffee is
Tcoffee = Tf 2 Tcoffee,i = 84.9°C 2 95.0°C = 210.1°C = 210.1 K
The heat that flows into the coffee is
Qcoffee = mcoffeecwater Tcoffee
= (0.350 kg)(4186 J kg21 # K21)(210.1 K)
= 21.48 * 104 J
This is negative because heat flows out of the coffee.