Transcript Slide 1

Try #s 1 and 2 on your own before we begin!
1. Q = mcT
= (432g)(4.18J/goC)(18.0oC-71.0oC)
= -95705.28 J or
Q
2. c =
m T
=
-95.7 kJ
382 J
(66.7 g)(92.5oC - 1.01oC)
= 0.0625 J/goC
+
Metal Cylinder at 100oC
Water at 20.0o C
Hot metal cylinder will heat up the water.
-Q = the hot item
Q = the cooler item
The final temperature will
be the same for BOTH the
metal and the water.
3. A piece of metal, mass of 45.5 g and a temperature of
80.5oC is dropped into 192 g of water at 15.0oC. The final
temperature of the system is 15.8oC. What is the specific heat
of the metal? (Hint: -Q=Q)
-Qmetal = Qwater
- mc T = mc T
-(45.5g) c (15.8oC - 80.5oC) = (192 g)(4.18J/goC)(15.8oC - 15.0oC)
-(45.5g) c (-64.7oC) = 642.048
2943.85 c = 642.048
2943.85 = 2943.85
c = 642.048
2943.85
c = 0.218 J/goC
4. A piece of unknown metal with a mass of 14.9g is heated to
100.0oC and dropped into 75.0g of 20.00 C water. The final
temperature of the system is 28.5oC. What is the specific heat
of the metal?
-Qmetal = Qwater
- mc T = mc T
-(14.9g) c (28.5oC - 100oC) = (75.0 g)(4.18J/goC)(28.5oC - 20.0oC)
-(14.9g) c (-71.5oC) = 2664.75
1065.35 c = 2664.75
1065.35 = 1065.35
c = 2664.75
1065.35
c = 2.50 J/goC
6. Suppose a piece of gold (specific heat of 0.128J/goC) with a
mass of 21.5 g at a temperature of 95.00oC is dropped into an
insulated calorimeter containing 125.0 g of water at 22.00oC.
What is the final temperature of the system?
(Hint: Heat gained must equal heat lost.)
-Qmetal = Qwater
- mc T = mc T
- (21.5g)(0.128J/goC) (Tf - 95.00oC) = (125.0g)(4.18J/goC)(Tf - 22.00oC)
-2.752
(Tf - 95.00oC)
=
-2.752 Tf + 261.44
+2.752 Tf + 11495
11756.44
525.252
522.5 (Tf - 22.00oC)
= 522.5 Tf
+2.752 Tf
=
525.252 Tf
525.252
- 11495
+ 11495
Tf = 22.4oC