Thermo Chemistry Review Problems
Name:___________________
Period:______
q = m  t  C
The specific heat of ice is 2.1 J/g  oC
Heat of fusion for ice to water is 334 J/g
The specific heat of water is 4.184 J/g  oC
Specific Heat Sample Questions
1. Calculate the amount of heat needed to increase the temperature of 250g of water from 20oC to 46oC.
q = m  t  C
q = 250g  26oC  4.18J/g  oC
q = 37,620J or 37.6kJ
2. Calculate the specific heat of copper given that 204.75J of energy raises the temperature of 15g of copper
from 25o to 60o
q = m  t  C
C= q/m  t
C = 204.75J /(15g  35oC)
C= 0.39 J/goC
3. A total of 216 J of energy is required to raise the temperature of aluminum from 15 o to 35oC. Calculate the
mass of aluminum. (Specific Heat of aluminum is 0.90 J/g  oC)
q = m  t  C
m= q/C  t
m= 216J / (0.90 J/goC  20oC )
m= 12g
4. The initial temperature of 150g of ethanol was 22oC. What will be the final temperature of the ethanol if
3240 J was needed to raise the temperature of the ethanol? (Specific heat of ethanol is 2.44 J/g  oC)
t = q/m  C
t = 3240J/(150g  2.44J/goC)
t = 8.85oC
Tfinal = 22oC + 8.85oC = 30.9oC
Heat of Fusion Sample Questions
5. How much energy is required to melt a 52.5g block of ice at 0 oC and convert it to 52.5g of water at 0 oC?
Express your answer in kJ. Heat of fusion for ice to water is 334 J/g
q = m  Hf
q = 52.5g  334j/g = 17535J or 17.535kJ
6. Calculate the total energy required to convert a 96g block of ice at -24.0 oC to water at 28.0 oC.
Hint: this is a four step process.
The specific heat of ice is 2.1 J/g  oC
Heat of fusion for ice to water is 334 J/g
The specific heat of water is 4.184 J/g  oC
Step 1 - warm ice from -24 oC to 0 oC
q = m  t  C
q = 96g  (0 oC - (-24 oC)  2.1j/g = 4838.4J or 4.838kJ
Step 2 - convert ice at 0 oC to water at 0 oC (melt the ice)
q = m  Hf
q = 96g  334j/g = 32064J or 32.064kJ
Step 3 - heat water from 0oC to 28 oC
q = m  t  C
q = 96g  (28 oC - 0 oC)  4.184j/g = 11246.592J or 11.24659kJ
Step 4 - sum all energies
4.838kJ + 32.064kJ + 11.24659kJ = 48.1kJ
7. Calculate the total energy required to convert a 2.7g block of ice at -17.0 oC to water at 42.0 oC.
Hint: this is a four step process.
The specific heat of ice is 2.1 J/g  oC
Heat of fusion for ice to water is 334 J/g
The specific heat of water is 4.184 J/g  oC
Step 1 - warm ice from -17 oC to 0 oC
q = m  t  C
q = 2.7g  (0 oC - (-17 oC)  2.1j/g = 96.39J
Step 2 - convert ice at 0 oC to water at 0 oC (melt the ice)
q = m  Hf
q = 2.7g  334j/g = 901.8J
Step 3 - heat water from 0oC to 42 oC
q = m  t  C
q = 2.7g  (42 oC - 0 oC)  4.184j/g = 474.4656J
Step 4 - sum all energies
96.39J + 901.8J + 474.4656J = 1472.7J or 1.5kJ
8. Calculate the total energy released when 100g of water at 50.0 oC is cooled to ice at -30.0 oC.
Hint: this is a four step process.
The specific heat of ice is 2.1 J/g  oC
Heat of fusion for ice to water is 334 J/g
The specific heat of water is 4.184 J/g  oC
Step 1 - cool water from 50 oC to 0 oC
q = m  t  C
q = 100g  -50 oC  4.184j/g = -20920J or -20.9kj
Step 2 - convert water at 0 oC to ice at 0 oC (freeze the water to ice)
q = m  Hf
q = 100g  -334j/g = -33400J or -33.4kJ
Step 3 - cool ice from 0oC to -30 oC
q = m  t  C
q = 100g  (-30oC)  2.1j/g = -6300J or -6.3kJ
Step 4 - sum all energies
-20.9kJ - 33.4kJ - 6.3kJ = -60.6kJ
CHALLENGE PROBLEM!!!
9. Calculate the total energy required to convert 1.0 mol of H20 from -25.0 oC to 125.0 oC.
Hint: this is a six step process!
The specific heat of ice is 2.1 J/g  oC
Heat of fusion for ice to water is 334 J/g
The specific heat of water is 4.184 J/g  oC
Heat of vaporization of water to steam is 2256.9 J/g
The specific heat of steam is 1.84 J/g  oC
Step 1 - warm ice from -25 oC to 0 oC
1 mol H20 = 18.02 g H20
q = m  t  C
q = 18.02 g  (25 oC)  2.1j/g = 940J or 0.94kJ
Step 2 - convert ice at 0 oC to water at 0 oC (melt the ice)
q = m  Hf
q = 18.02 g  334j/g = 6018.68J or 6.01kJ
Step 3 - heat water from 0oC to 100 oC
q = m  t  C
q = 18.02 g  100 oC  4.184j/g = 7539.568J or 7.54kJ
Step 4 - convert water at 100 oC to steam at 0 oC (vaporize the water)
q = m  Hf
q = 18.02 g  2256.9j/g = 40669.338J or 40.7kJ
Step 5 - heat steam from 100oC to 125 oC
q = m  t  C
q = 18.02 g  25oC  1.84j/g = 828.92J or 0.83kJ
Step 6 - sum all energies
0.94kJ + 6.01kJ + 7.54kJ + 40.7kJ + 0.83kJ = 56.02kJ