The ceramic coffee cup of 116g in Figure 16–16, with c=1090J/kg.K and is
initially at room temperature (24.0 °C). If 225 g of 80.3 °C coffee and 12.2
g of 5.00 °C cream are added to the cup, what is the equilibrium
temperature of the system? Assume that no heat is exchanged with the
surroundings, and that the specific heat of coffee and cream are the same
as the specific heat of water.
[Answer]
Step 1: define variables
Mceramic = 0.116kg; Mcream = 0.0122kg; Mcoffee =0.225kg
Tceramic,initial = 24.0 °C; Tcream,initial =5°C; Tcoffee,initial =80.3°C
Looking for Tfinal
Step 2: find the corresponding physical model and state applicable theory
According to calorimetrical theory,
Qceramic + Qcream + Qcoffee = 0
Step 3: Solve the equations
Qcup  Qcof  Qcrm  0
0  mcup ccup (T  Tcup )  mcof cw (T  Tcof )  mcrm cw (T  Tcrm )
 T [mcup ccup  (mcof  mcrm )cw ]  [mcup ccupTcup  (mcof Tcof  mcrmTcrm )cw ]
mcup ccupTcup  (mcof Tcof  mcrmTcrm )cw
T
mcup ccup  (mcof  mcrm )cw




(0.116 kg) 1090 kgJ K (24.0 C)  [(0.225 kg)(80.3 C)  (0.0122 kg)(5.00 C)] 4186 kgJ K
 70.5 C



(0.116 kg) 1090 kgJ K  (0.225 kg  0.0122 kg) 4186 kgJ K

