Three Equations Needed

Go to the Time-Temperature Graph file

Problems using one part of the T-T graph

Problems using two parts of the T-T graph

Problems using four parts of the T-T graph

Problems using five parts of the T-T graph

Return to Thermochemistry Menu

**Problem #1:** A 36.0 g sample of water is initially at 10.0 °C. How much energy is required to turn it into steam at 200.0 °C? (This example starts with a temperature change, then a phase change followed by another temperature change.)

**Solution:**

q = (36.0 g) (90.0 °C) (4.184 J g¯^{1}°C¯^{1}) = 13,556 J = 13.556 kJq = (40.7 kJ/mol) (36.0 g / 18.0 g/mol) = 81.4 kJ

q = (36.0 g) (100.0 °C) (2.02 J g¯

^{1}°C¯^{1}) = 7272 J = 7.272 kJq = 102 kJ (rounded to the appropriate number of significant figures)

**Problem #2:** A 72.0 g sample of ice is at 0 °C. How much energy is required to convert it to steam at 100.0 °C? (This example begins with phase change, then a temperature change and then a second phase change, areas two, three and four on the time-temperature graph.)

**Solution:**

q = (6.02 kJ/mol) (72.0 g / 18.0 g/mol) = 24.08 kJq = (72.0 g) (100.0 °C) (4.184 J g¯

^{1}°C¯^{1}) = 30,125 J = 30.125 kJq = (40.7 kJ/mol) (72.0 g / 18.0 g/mol) = 162.8 kJ

q = 217 kJ (rounded to the appropriate number of significant figures)

**Problem #3:** Calculate the heat released by cooling 54.0 g H_{2}O from 57.0 °C to minus 3.0 °C. (Cools from 57.0 to zero, a phase change, the cools from zero to -3; three equations needed.)

**Solution:**

q = (54.0 g) (57.0 °C) (4.184 J g¯^{1}°C¯^{1}) = 12,878 J = 12.878 kJq = (6.02 kJ/mol) (54.0 g / 18.0 g/mol) = 18.06 kJ

q = (54.0 g) (3.0 °C) (2.06 J g¯

^{1}°C¯^{1}) = 333.73 J = 0.334 kJq = 31.3 kJ (rounded to the appropriate number of significant figures)

**Problem #4:** How much energy is required to heat 125.0 g of water from -15.0 °C to 35.0 °C?

The solution:

q = (125.0 g) (15.0 °C) (2.06 J g¯

^{1}°C¯^{1}) = 3862.5 J = 3.8625 kJq = (6.02 kJ/mol) (125.0 g / 18.0 g/mol) = 41.806 kJ

q = (125.0 g) (35.0 °C) (4.184 J g¯

^{1}°C¯^{1}) = 18,305 J = 18.305 kJq = 64.0 kJ (rounded to the appropriate number of significant figures)

**Problem #5:** How much energy in kJ is needed to heat 5.00 g of ice from -10.0 °C to 30.0 °C? The heat of fusion of water is 6.02 kJ/mol, and the molar heat capacity is 36.6 J/mol K for ice and 75.3 J/mol K for liquid water

**Solution:**

1) The ice will do three things:

a) heat up from -10.0 °C to 0 °C

b) melt at 0 °C

c) warm up (as a liquid) from 0 to 30.0 °C

2) Each one of those steps requires a calculation. Then, the results of the three calcs will be added together for the final answer.

q_{a}= (5.00 g / 18.0 g /mol) (10 °C) (36.6 J/mol K)

q_{b}= (5.00 g / 18.0 g /mol) (6.02 kJ/mol)

q_{c}= (5.00 g / 18.0 g /mol) (30 °C) (75.3 J/mol K)q

_{a}and q_{c}will give J for an answer while q_{b}gives kJ. Convert q_{a}and q_{c}to kJ before adding.The Celsius and the Kelvin will cancel in the q

_{a}and q_{c}equations. This is because the size of one °C equals the size of 1 K. The Celsius values in q_{a}and q_{c}are differences, not specific temperatures. You can confirm this by converting -10 and 0 to their Kelvin values and then subtracting. You will get 10 K.

2) The answer:

0.10167 kJ + 1.67 kJ + 0.62750 kJ = 2.40 kJ (to three sig figs)

**Problem #6:** A 10.35 kg block of ice has a temperature of -22.3 °C. The block absorbs 4.696 x 10^{6} J of heat. What is the final temperature of the liquid water?

**Solution:**

1) Raise the temperature of the ice to 0 °C:

q = (10350 g) (22.3 °C) (2.06 J/g¯^{1}°C¯^{1})q = 475458.3 J

2) Melt the ice at 0 °C:

q = (6020 J/mol) (10350 g / 18.015 g/mol)q = 3458617.818 J

3) How many Joules remain?

total used to this point = 3.934 x 10^{6}Jtotal remaining: 4.696 x 10

^{6}J minus 3.934 x 10^{6}J = 7.62 x 10^{5}J

4) Determine temperature increase of liquid water:

7.62 x 10^{5}J = (10350 g) (x) (4.184 J/g¯^{1}°C¯^{1})x = 1.76 °C

**Problem #7:** How many grams of ice at -5.6 °C can be completely converted to liquid at 10.1 °C, if the available heat for this process is 4.74 x 10^{3} kJ?

**Solution:**

1) Raise the temperature of the ice to zero Celsius:

q = (x) (5.6 °C) (2.06 J/g¯^{1}°C¯^{1})

2) Melt the ice at zero Celsius:

q = (6020 J/mol) (x / 18.015 g/mol)

3) Raise the temperature of the ice to 10.1 Celsius:

q = (x) (10.1 °C) (4.184 J/g¯^{1}°C¯^{1})

4) 4.74 x 10^{6} J are used in total:

4.74 x 10^{6}J = [(x) (5.6 °C) (2.06 J/g¯^{1}°C¯^{1})] + [(6020 J/mol) (x / 18.015 g/mol)] + [(x) (10.1 °C) (4.184 J/g¯^{1}°C¯^{1})]The solution for x is left to the reader.

**Problem #8:** Bromine melts at -7.25 °C and boils at 58.8 °C. The enthalpy of fusion of bromine is 10.57 kJ/mol and the enthalpy of vaporization of bromine is 29.96 kJ/mol. The specific heat of liquid bromine is 0.474 J/g**⋅**K. How much heat, in kJ, is required to convert 25.0 g of solid bromine at -7.25 °C to the gas phase at 58.8 °C?

**Solution:**

1) Three calculations are needed:

a) melt bromine at -7.25

b) heat liquid bromine from -7.25 to 58.8

c) boil bromine at 58.8

2) Here are the calculation set-ups:

q_{a}= (25.0 g / 159.808 g/mol) (10.57 kJ/mol)

q_{b}= (25.0 g) (66.05 K) (0.474 J/g⋅K)

q_{c}= (25.0 g / 159.808 g/mol) (29.96 kJ/mol)159.808 g/mol is the molar mass of Br

_{2}.

66.05 K is the absolute temperature difference between -7.25 °C and 58.8 °C.

Note that I used it in Kelvin, rather than Celsius.

3) Convert the q_{b} answer to kJ (from J) and then add all three answers. Round off as appropriate and you're done.

Go to the Time-Temperature Graph file

Problems using one part of the T-T graph

Problems using two parts of the T-T graph

Problems using four parts of the T-T graph