Ladue High School Chemistry
Ch. 14 - Solids and Liquids
Energy Calculations Homework #2 - Practice
sice = 2.06 J/g •˚C
sliquid = 4.184 J/g •˚C
ssteam = 2.02 J/g •˚C
∆Hfus= 334 J/g
∆Hvap= 2260 J/g
density = 1 g/mL
1. How much heat is required to boil and vaporize 1 cup of water (250 mL) which is
already at 100˚C?
Phase change = use ∆Hvap
250.0 mL (1 g/mL) = 250.0 g H2O
250.0 g (2260 J/g) = 565000 J (1 kJ/1000J) = 565 kJ
2. How much energy must be removed from 523.0 mL of water at 85.0 ˚C to lower
the temperature to 21.0˚C?
No phase change, so use q = ms∆T
523 mL (1 g/mL) = 523 g
∆T = -64˚C
q = (523 g)(4.184 J/g•˚C)(-64˚C)
q = -140046 J (1 kJ/1000 J) = -140 kJ
3. An ice tray contains 811.0 g of water at 38.0 ˚C. How much energy must be
removed to produce ice at 0.0˚C?
2 steps: Temp change (use q = ms∆T, using specific heat of liquid
water), and phase change (use q = m∆Hfus)
1st step - temp change:
∆T = -38˚C
m = 811.0 g
s = 4.184 J/g˚C
q = (811.0g)(4.184 J/g˚C)(-38 ˚C)
q = -129000 J (1 kJ/1000 J) = -129 kJ
2nd step - phase change (exothermic since going from higher energy
state to lower energy state):
(811.0 g)(-334 J/g) = -271,000 J (1 kJ/1000 J) = -271 kJ
Total Energy = sum of energy of the 2 steps
-129 kJ + -271 kJ = -400 kJ
4. How many kilojoules of heat energy are needed to change 49.6 g of ice at -15˚C
to water at 58˚C?
3 steps: temp change from -15˚C to 0.0 ˚C, phase change from
to liquid, temp change from 0.0 ˚C to 58˚C.
solid
Step 1 - temp change - use q = ms∆T, using specific heat of ice
m = 49.6 g
∆T = 15 ˚C
s = 2.06 J/g • ˚C
q = (49.6 g)(2.06 J/g•˚C)(15 ˚C)
q = 1500 J
Step 2 - phase change - use ∆Hfus
q = (49.6 g)(334 J/g)
q = 16600 J
Step 3 - temp change - use q = ms∆T, using specific heat of liquid
water
∆T = 58 ˚C
q = (49.6 g)(4.184 J/g ˚C)(58˚C)
q = 12000 J
Total energy = 1500 J + 16600 J + 12000 J = 30,100 J (1 kJ/1000 J) =
30.1kJ
5. How many kJ will be released as 575.0 g of steam at 176.0 ˚C are changed to ice
at -75 ˚C?
5 steps:
1st step = reduce temp of steam from 176˚C to 1OO˚C
2nd step = phase change - steam to liquid
3rd step = reduce temp of liquid from 100˚C to 0.00˚C
4th step = phase change - liquid to ice
5th step = reduce temp of ice from 0.00 ˚C to - 75.0 ˚C
1st step, use q = ms∆T, using specific heat of steam
∆T = - 76˚C
q = (575.0 g)(2.02 J/g•˚C)(-76˚C)
q = -88274 J
2nd step, use q = m∆Hvap (which will have a neg value, since energy being
released)
q = (575.0 g)(-2260 J/g)
q = -1299500 J
3rd step, use q = ms∆T, using specific heat of liquid water
∆T = - 100 ˚C
q = (575.0 g)(4.184 J/g • ˚C)(-100 ˚C)
q = -240580 J
4th step, use q = m∆Hfus (which will have a neg value, since energy
being released)
q = (575.0 g)(-334 J/g)
q = -192050 J
5th step, use q = ms∆T, using specific heat of ice (solid)
∆T = -75.0˚C
q = (575.0 g)(2.06 J/g • ˚C)(-75.0 ˚C)
q = -88837.5 J
Total energy =
-88274 J + -1299500 J + -240580 J + - 192050 J + -88837.5 J
= -1910000 J (1 kJ/1000 J) = -1910 kJ
6. Years ago, a block of ice with a s mass of 20.0 kg was used daily in a home
icebox. The temperature of the ice was 0.0˚C as delivered. As it melted, how
much heat, in kJ, did a block of ice of this size absorb?
No temp change, just phase change, so use ∆Hfus (pos value,
endothermic process)
(20.0kg)(1000 g/ 1 kg)(334 J/ g)(1 kJ/1000J) = 6680 kJ