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4. 10 points] At a crime scene, the forensic investigator notes that the 8.2-g lead bullet that
was stopped in a door-frame melted completely on impact. Assume the bullet was red at
room temperature (20C).
The heat of fusion for lead is 5.9 kcal/kg and its melting point is 327C.
The specic heat of lead is 0.031 cal/g*C or 130 J/kg C
(a) 3 points] What is the heat required to get to the melting point?
Heat = 8:2g (327 ; 20)C 0:031cal=gC = 78 cal
(b) 3 points] What is the heat required for melting the bullet?
Heat melting = 8:2g 5:9 103 10;3 = 48.4 cal
(c) 2 points] What is the energy in joules required to heat and melt the bullet?
Energy = 4:18 (78 + 48:4) j = 528 joules
(d) 2 points] What does the investigator calculate was the minimum muzzle velocity of the
gun?
KE = 21 mv2 = 528 joules.
p
v = 2 528=8:2 10;3 = 359m=s
1
5. 15 points] You plan to have iced tea for your afternoon party. You have an insulated
container with 3 kg of tea (essentially water) at 20C, to which you add ice at -10C.
(a) 5 points] How much ice at -10C do you need to add to the 20C tea in order to have a
resulting mixture of tea (water) and 0.2 kg of ice in equilibrium? What is the equilibrium
temperature?
The equilibrium temperature will be 0 since we have ice and water together. The ice will
warm and all but 0.2 kg will melt. (We will assume the specic heat capacity of ice to be
the same as water in reality it is about half.)
The heat lost by the water is
Qw = mw cw Tw = 3000g 1cal=gC ;20 C = ;6 104cal
The heat gained by the ice is
Qice = miceciceTice + (mice ; 200g)Lf
= mice 1cal=gC 10C + (mice ; 200g) 80cal=g
= (10 + 80)micecal=g ; 1:6 104cal
The two heats have to sum to zero so we have (10 + 80)micecal=g ; 1:6 104cal = 6 104cal
or 90mice = 7:6 104g yielding mice = 844g = 0:844kg
(b) 5 points] Suppose the container of the iced tea is left in the state equilibrium describe in
part (a) and the temperature outside the container is xed at 20C. Suppose further that the
container has a total surface area of 2500 cm2, a thickness of 1 cm and is made of material
with thermal conductivity 0.025 W/mC. While there is still ice left, nd the rate at which
heat energy is entering the container and how long it takes the ice to melt.
We use Newton's heat conductivity equation
dQ = ;A T = 0:025W=mC 2500cm2 20C = 0:025 25 20W = 12:5W = 3cal=s
dt
l
1cm
At the end of part (a), we have 0:2 kg of ice left in our mix. The heat needed to melt that
ice is Q = miceLf = 200g 80cal=g = 16000cal. At three calories per second that takes a
time t = 16000=3s = 5350s = 89:2min = 1:49h.
(c) 5 points] What was the entropy change in part (b) and in part (a)?
The entropy change in part (b) is very easy to calculate. For the ice water:
S = Q=T = 16000cal=273K = 58:6 cal=K
1
and for the surroundings
S = Q=T = ;16000cal=293K = ;54:6 cal=K
Then the net result is S = 4:0 cal=K:
The entropy change
in part (a) is much more complicated to calculate. In general, dS =
R
dQ=T or S = dQ=T .
For the water
S =
Z
273K
293K
mw cw dT
T = mw cw ln(273=293) = 3000g 1cal=gK ln(273=293) = ;212 cal=K
For the ice
S
= micecice ln(Tf =Tice) + (mice ; 200g)Lf =Tf
= 844 1cal=K ln(273=263) + (844 ; 200) 80cal=273K = 31:5 + 189 cal=K
= 220 cal=K
The total is then S = 8 cal=K.
Note in both cases that S > 0 which is absolutely required since we are performing an
irreversible process: letting heat travel from a hot object to a cold one.
2