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Example Problem #1: A cubic block of uranium metal (specific heat = 0.117 J g¯1 °C¯1) at 200.0 °C is dropped into 1.00 L of deuterium oxide of "heavy water" (specific heat = 4.211 J g¯1 °C¯1) at 25.5 °C. The final temperature of the uranium and the deuterium oxide mixture is 28.5 °C. Given the densities of uranium (19.05 g/cm3) and deuterium oxide (1.11 g/mL) what is the length of the side of the uranium cube?
1) Determine the mass of uranium dropped into the D2O. We assume the heat lost by the uranium equals the heat gained by the water:
(x) (174.5) (0.117) = (1110 g) (3.0) (4.211)
x = 687.8283 g (I kept some guard digits)
2) Determine the volume of the uranium cube:
density = mass / volume
19.05 = 687.8283 / x
x = 36.053979 cm3
3) Determine length of side of cube:
[cube root]36.053979 cm3 = 3.30 cm (to three sig figs)
Example Problem #2: 175.0 g pure H2O was placed in a constant-pressure calorimeter and chilled to 10.0 °C. 9.80 g pure H2SO4 (also at 10.0 °C) was added, stirred and the temperature rose to 19.8 °C.
(a) What mass increased in temperature?
(b) What was Δt?
(c) What was the chemical reaction?
(d) What do we assume about the specific heat?
(e) Calculate the energy change for the reaction.
(f) What state function does this energy change represent?
(g) Calculate the energy change in kJ/mole.
(a) Both water AND sulfuric acid started at 10 and went to 19.8, so let's add 175.0 + 9.80 to get 184.80 g. That's the mass that increased in temperature.
(b) The Δt equals 19.8 minus 10.0 for 9.8 °C.
(c) This answer may be a bit confusing, until you get to the acid base section of your studies. Try not to worry about it too much.
H2SO4 + 2 H2O ---> 2 H3O+ + SO42¯
The protons on the sulfuric acid (an acid is a proton donor) were transferred to the water (called a proton acceptor; it is acting as a base). For those who have not hit acid base yet in class, this is an understanding of the reaction in Brønsted-Lowry terms. This particular acid base theory is quite important. Give it some good study when you get there.
(d) We are going to assume the specific heat of liquid water and sulfuric acids are the same. The truth is that they are not, but there is a small amount of acid compared to the amount of water, so the assumption introduces an amount of error that is tolerable.
Of course, the question is when does it become intolerable. That really is a question for another time, because here at the beginning level it's OK to make some assumptions that later on will get thrown out.
(e) The symbolic equation we will use is:
q = (mass) (Δt) (Cp)
Putting numbers into the equation:
x = (184.80 g) (9.8 °C) (4.18 J g¯1 °C¯1)
(f) Enthalpy. Remember the definition: heat flow at constant pressure. Also, the term "state function" is used. Be assured "state function" is a very, very important idea. Enthalpy change, ΔH, the change in the Gibbs free energy, ΔG, and entropy change, ΔS, are all state functions. Work (w) and heat (q) ARE NOT state functions.
(g) This question is about kJ/mol of sulfuric acid, NOT water. This is almost always the case, that we are interested in what happens to the substance put into the water. So, we use the 7570 J answer from (e) as 7.57 kJ and divide by the moles of sulfuric acid:
7.57 kJ / 0.10 mol = 75.7 kJ mol¯1
If you're unsure how I got 0.10 mol of H2SO4, than ask a friend for help.
Example Problem #3: A 4.40 g lead bullet moving at 250.0 m/s strikes a steel plate and stops. If all the bullet's kinetic energy is converted to thermal energy (and none is transferred to the steel plate), what is the bullet's temperature change?
1) The kinetic energy (measured in Joules) is calculated using the following equation:
K.E. = (1/2) mv2
where m is in kilograms and v is in meters per second. This will give the unit kg m2 s-2, which are the units that a Joule is measured in. Keep in mind, Joule is the name of the unit of energy; the unit it is measured in is kg m2 s-2.
K.E. = (1/2) (0.00440 kg) (250.0 m/s)2 = 275 J
Comment: the kinetic energy equation is not normally taught in a thermochemistry setting.
2) We next turn to the heat capacity equation:
q = (mass) Δt) (Cp)
to continue solving this problem. However, the problem does not provide a necessary value, namely the specific heat (symbol = Cp) for solid lead. Turning to the Internet, we find this value to be 0.128 J g-1 °C-1.
Substituting, we now find:
275 J = (4.40 g) (x) (0.128 J g-1 °C-1)
x = 488 °C
Note that I used 4.40 g rather than 0.00440 kg.
Example Problem #4: A typical microwave oven uses radiation with a 12.2 cm wavelength.
a) How many moles of photons of this radiation are required to raise the temperature of 405.0 g of water from 26.5 °C to 99.8 °C?
b) The watt is a unit of power - the rate at which energy is delivered or consumed: 1 W = 1 J/s. Assume that all the energy of a 760.0 W microwave oven is delivered to the heating of water described in (a). How long will it take to heat the water?
Solution to part a:
1) determine the Joules of energy necessary to heat the water:
q = (mass) Δt) (Cp)2) Using λν = c and E = hν, determine how many Joules are carried by one photon of wavelength 12.2 cm:
q = (405.0 g) (73.3 °C) (4.184 J g-1 °C-1)
q = 124208.316 J (I'll keep all digits and round off at the end.)
(12.2 cm) (ν) = 3.00 x 1010 cm s-1
ν = 2.459 x 109 s-1
E = (6.626 x 10-34 J s) (2.459 x 109 s-1)
E = 1.629 x 10-24 J per photon
3) Determine moles of photons required:
124208.316 J / 1.629 x 10-24 J per photon = 7.6232 x 1028 photons
7.6232 x 1026 photons / 6.022 x 1023 photons mol-1 = 1.266 x 105 moles of photons
Solution to part b:
1) The microwave delivers energy at the rate of 760.0 J/s.
2) 124208.316 J / 760.0 J/s = 163.4 seconds (a bit more than 2.5 minutes)
Example Problem #5: Citric acid has three dissociable hydrogens. When 5.00 mL of 0.64 M citric acid and 45.00 mL of 0.77 M NaOH at an initial temperature of 26.0 °C are mixed, the temperature rises to 27.9 °C as the citric acid is neutralized. The combined mixture has a mass of 51.6 g and a specific heat of 4.0 J/g °C. Assuming no heat is transferred to the surroundings, calculate the enthalpy change for the reaction of 2.15 mols of citric acid in kJ. Two sig. figs. for the answer.
1) Determine J absorbed by the water:
q = (51.6 g (1.9 °C) (4.0 J / g °C) = 392.16 J
2) Determine moles of citric acid:
(0.0050 L) (0.64 mol / L) = 0.0032 mol
3) Determine moles of NaOH (just to show that it is in excess):
(0.045 L) (0.77 mol / L) = 0.03465 mol
The NaOH is very much in excess, as only 0.0096 mol of the NaOH is required to react with the three dissociable hydrogens.
4) Determine kJ/mol for citric acid:
392.16 J / 0.0032 mol = 122.55 kJ/mol
5) Determine enthalpy change for 2.15 moles of citric acid:
122.55 kJ/mol x 2.15 mol = 263.5 kJ
To two sig figs, the answer is 260 kJ (If your teacher insists, make it be -260 kJ, since the reaction was exothermic.).
Example Problem #6: The specific heat capacity of water is 4.184 J/g-°C. A 400.0 g sample of liquid water is exposed to the light emitted by a CO2 laser. The wavelength of the laser light is 1.064 x 104 nm. Assuming that all of the light energy is converted into heat, calculate the number moles of photons required to raise the temperature of the water by 5.00 °C.
1) Calculate Joules required to heat water:
q = (400.0 g) (5.00 °C) (4.184 J/g-°C)
q = 41840 J
2) Using Eλ = hc, calculate energy of one photon from the laser:
1.064 x 104 nm = 1.064 x 10¯5 m
E = [ (6.626 x 10¯34 J s) (3.00 x 108 m s¯1) ] / 1.064 x 10¯5 m
E = 1.868233 x 10¯20 J
3) Calculate photons required to supply 41480 J:
41840 J / 1.868233 x 10¯20 J = 2.23955 x 1024 photons
4) Convert to moles of photons:
2.23955 x 1024 / 6.022 x 1023 = 3.72 moles
Example Problem #7: An ice cube whose mass is 0.080 kg is taken from a freezer where its temperature was -12.0 °C and dropped into a glass of water at 0.0 °C. If no heat is gained or lost from outside, how much water freezes onto the cube? The latent heat of fusion of water is 33.5 x 104 J/kg and the specific heat capacity of ice is 2000 J/kg °C.
1) Calculate amount of heat to raise ice to zero Celsius:
q = (0.080 kg) (12.0 °C) (2000 J/kg °C) = 1920 J
2) Calculate amount of water that freezes when 1920 J is removed:
1920 J = (x) (33.5 x 104 J/kg)
x = 5.73 x 10¯3 kg = 5.73 g
Example Problem #8: The latent heat of fusion of water is 335,000 J/kg & the specific heat capacity of ice is 2,000 J/(kg C). If I introduce 2,000 J of heat to a melting block of ice with mass 1.0 kg at 0 degrees C, the final temperature is equal to:
a. 0 degrees C
b. 2 degrees C
c. 40 degrees C
d. 2,000 degrees C
You need to add 335,000 J to get the entire 1.0 kg of ice melted. Adding 2000 J only manages to melt a small portion of the ice. Therefore the final temperature is still 0.0 deg C. Most of it is ice at zero and a small amount of it is liquid at zero.
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