Name: _____________________________________ Date: ______________ Section: ___________________________
THERMODYNAMICS: 6.4 QUIZ
1) Equal mass samples of aluminum and lead absorb the same amount of heat energy. Which sample experiences
the greatest change in temperature? Justify your answer.
Lead. Lower specific heat means that it will change temperature more readily. If both aluminum and lead absorb
the same amount of heat, lead will have the greatest change in temperature.
2) A 5.00 kg casserole dish released 96,600 J of heat as it cooled from 84.0°C to 61.0°C. Determine the material the
dish was made from.
𝑄 = 𝑚𝑐𝑝 (𝑇𝑓 − 𝑇𝑖 )
−96,000 = 5000𝑐𝑝 (61 − 84)
−96,000 = −150,000𝑐𝑝
𝑐𝑝 = 0.84 𝐽/𝑔°𝐶
Material = GLASS
Q = -96,000 J
m = 5000 g
cp = ?
Ti = 84 C
Tf = 61 C
Temperature (degrees Celcius)
3) The heating curve below illustrates the temperature of a sample of 0.4 kg of Raquelenium as a function of time.
Assume heat was absorbed at a constant rate of 1.0 kilojoule per minute.
60
50
40
30
20
10
0
0
2
4
6
8
10
12
14
Time (min)
a) Rank the phases of matter of Raquelenium in order of decreasing specific heat capacity.
Least steep  Most steep
Gas  Liquid  Solid
b) Determine the latent heat of vaporization for Raquelenium.
Q = 10kJ – 6kJ = 4kJ = 4000 J
m = 0.4 kg = 400 g
Q = mL
4000 = 400L
Lv = 10 J/g
4) A 45 g copper penny that had been sitting out in the sun was placed into 50 g of water at 20°C to cool it down.
The two were allowed to sit and reach thermal equilibrium.
a. Circle the correct answer for each of the bolded choices in parentheses. According to the Zeroth Law of
Thermodynamics, heat flows from (the penny / the water) to (the penny / the water) until the two
reach thermal equilibrium. [Heat flows from high temperature  Low temperature until the two reach
equilibrium]
b. If the final temperature of the water is 24°C, determine the initial temperature of the penny. Assume
heat losses to the environment are negligible.
PENNY
m = 45 g
cp = 0.385 J/gC
Ti = ?
Tf = 24 C
− (𝑚𝑐𝑝 (𝑇𝑓 − 𝑇𝑖 ))
𝑝𝑒𝑛𝑛𝑦
WATER
m = 50 g
cp = 4.184 J/gC
Ti = 20 C
Tf = 24 C
= (𝑚𝑐𝑝 (𝑇𝑓 − 𝑇𝑖 )
𝑤𝑎𝑡𝑒𝑟
−(45(0.385)(24 − 𝑇𝑖 )) = (50(4.184)(24 − 20))
17.325(24 − 𝑇𝑖 ) =
−415.8 + 17.325𝑇𝑖 = 836.8
17.325𝑇𝑖 = 1252.6
𝑇𝑖 = 72.3°C
5) During the process of vinegar distillation, pure acetic acid must first be vaporized and then infused into water.
To do this, 183,000 J of energy are used to heat an unknown quantity of acetic acid from 31°C in the liquid phase
to 130°C in the vapor phase. The thermodynamic properties of acetic acid are listed in the table below. Using
this information, determine the mass of the acetic acid sample.
Specific Heat Capacities (J/g°C)
Solid
Liquid
Gas
1.569
2.043
Melting Point
(°C)
Latent Heat of
Fusion (J/g)
Boiling Point
(°C)
17
196
118
1.147
Temperature Change (Liquid)
m=?
cp = 2.043 J/gC
Ti = 31 C
Tf = 118 C
Temperature Change (Gas)
m=?
cp = 1.147 J/gC
Ti = 118 C
Tf = 130 C
Phase Change (l  g)
m=?
L = 395 J/g
Q = m(2.043)(118 – 31)
Q = m(1.147)(130 – 118)
Q = m(395)
177.7m
TOTAL Q = 177.7m + 13.76m + 395m
183,000 = 177.7m + 13.76m + 395m
183,000 = 586.5m
m = 312 g
13.76m
395m
Latent Heat
of
Vaporization
(J/g)
395