C hanges
in
S tate P roblems
1. How much energy is needed to change a 25.0 g ice cube at
-10.0 °C to steam at 110.0°C? The
specific heat of steam is 0.50 cal/g°C.
2. An 850.0 mg sample of lead is kept in a
basement at 10.0 °C . If 25.0 cal of heat
energy are supplied to it from a nearby
furnace, will the sample melt
completely? Lead melts at 327°C.
3. 0.500 kcal of heat energy are supplied to a 50.00 g block of lead at
room temperature (20.0°C). What temperature and state will the
lead reach? Lead melts at 327°C and boils at 1750°C.
4. A 2.50 kg sample of molten silver at its melting point of 961°C is
placed in a container of liquid nitrogen (-195°C). How much
energy must be removed to bring the solid silver down to the
temperature of the N2 (l)?
5. Suppose there is a beaker containing 25.0 mL of
methanol in front of you in the lab, which is at
about 20°C. If you add 500 cal of heat energy, will
it
boil? The bp of methanol is 65°C. Assume 60% of the energy is
“wasted” in heating the beaker, and only 40% goes into heating the
methanol. The density of methanol is 0.80 g/mL, and its formula
is CH3OH.
Heat of Fusion
(kcal/mol)
Lead
Methanol
Silver
Water
1.14
0.765
2.84
1.44
Heat of Vap.
Specific Heat
Specific Heat
(kcal/mol)
Solid (cal/g-°C)
Liquid (cal/g-°C)
42.5
8.01
61.7
9.7
0.031
--0.61
0.50
--0.057
--1.00
ANSWERS:
(1)
warm ice = (0.50 cal/g-°C)(25.0 g)(10°C) = 125 cal = 0.125 kcal
melt ice = (25.0 g)(1 mol/18.0 g)(1.44 kcal/1 mol) = 2.00 kcal
heat water = (1.00 cal/g-°C)(25.0 g)(100°C) = 2500 cal = 2.50 kcal
boil water = (25.0 g)(1 mol/18.0 g)(9.7 kcal/1 mol) = 13.5 kcal
heat steam = (0.50 cal/g-°C)(25.0 g)(10°C) = 125 cal = 0.125 kcal
TOTAL = 18.3 kcal
(2)
heat solid Pb = (0.031 cal/g-°C)(0.850 g)(327 - 10°C) = 8.35 cal needed
melt Pb = (0.850 g)(1 mol/207.2 g)(1.14 kcal/1 mol) = 0.00468 kcal = 4.68 cal
total needed to melt Pb = 8.35 cal + 4.68 cal = 13.0 cal < 25.0 cal available
YES, the Pb will melt completely.
(3)
heat solid Pb = (0.031 cal/g-°C)(50.00 g)(327 - 20°C) = 476 cal = 0.476 kcal
With 0.500 kcal available, the lead will at least start to melt.
to melt Pb = (50.0 g)(1 mol/207.2 g)(1.14 kcal/1 mol) = 0.2755 kcal
This is much greater than the 0.500 kcal available, so …
The lead will be partially molten at 327°C.
A better answer is to determine just how much of the lead has melted.
There is 0.500 - 0.467 = 0.033 kcal available for melting the Pb. So …
(0.033 kcal)(1 mol/1,14 kcal)(207.2 g)/1 mol) = 6.00 g Pb can melt
There will be a mix of 6.00 g molten lead and 44.0 g solid lead
at 327°C.
(4)
freeze silver = (2500 g)(1 mol/107.9 g)(2.84 kcal/1 mol) = 65.8 kcal
cool solid = (0.61 cal/g-°C)(2500 g)[961-(-195°C)] = 1,763,000 cal = 1763 kcal
TOTAL = 1830 kcal
(5)
25.0 ml x (0.80 g/1 mL) = 20.0 g methanol
40% of 500 cal = 200 cal available to boil methanol
heat methanol = (0.057 cal/g-°C)(20.0 g)(65-20°C) = 51.3 cal
boil methanol = (20.0 g)(1 mol/32.0 g)(8.01 kcal/1 mol) = 5.01 kcal = 5010
cal
There is enough energy to start to boil the methanol,
but not enough to boil all of it, because 51.3 cal < 200 < 5010 cal.
Again, you can improve this answer by specifying how much of the
methanol will boil off. There is 200 - 39.9 = 160. calories available for
boiling. So …
(0.160 kcal)(1 mol/8.01 kcal)(32.0 g/1 mol) = 0.64 g. CH3OH boils off