Chapter 4 Problem Set.
2. You have a Fahrenheit thermometer in your left hand and a celcius thermometer in your
right hand. Both thermometers show the same reading in degrees. What is the
temperature?
Answer:
Relevant formulas:
Tc = 5/9 (Tf – 32)
Tf = 9/5*Tc + 32
You can use either formula and plug in Tc for Tf or vice versa.
Using the first formula and substituting Tc for Tf:
Tc = 5/9 (Tc – 32) 
Tc = 0.55*Tc – 17.8
.44*Tc = -17.6
Tc = -40.0C
Using the second formula and substituting Tf for Tc:
Tf = 9/5*Tf + 32
Tf = 1.8*Tf + 32
-.8Tf = 32
Tf = -40F
8. How much heat must be added to 1.5 kg of aluminum to raise its temperature from 20C to
200C?
Relevant formula:
Q = mcT [Heat transferred = (mass) X (specific heat) X (temperature change)].
We are given the mass of aluminum (1.5 kg), the temperature change (200C - 20C =
180C) and we can look up the specific heat of aluminum from Table 4-1 on pg 110 (0.92
kJ/kgC). Substituting into the formula:
Q = mcT = (1.5 kg) X (0.92 kJ/kgC) X (180C) = 248.4 kJ = 2.48 X 105 J (joules)
12. A 2.0-kg brass monkey has a volume of 240 cm3.
What is the density of the brass?
Relevant formula:
d = m/V (density = mass/volume)
We are given the mass of the brass monkey (2.0 kg = 2,000 g) and we are given the volume
of the brass monkey (240 cm3 ). Density is usually expressed as Kg/m3 or g/cm3. Since
we’ve already changed kilograms into grams:
d = m/V = (2000 g)/(240 cm3) = 8.33 g/cm3
24. The propellant gas that remains in an empty can of spray paint is at atmospheric
pressure. If such a can at 20C is thrown into a fire and is heated to 600C, how many times
atmospheric pressure is the new pressure inside the can? (The higher pressure may burst
the can, which is why such cans should not be thrown into a fire.)
The relevant formula here is:
(p1V1)/T1 = (p2V2)/T2
(pressure1 X volume1)/temperature1 = (pressure2 X volume2)/temperature2
We are given pressure 1 (atmospheric pressure [1 atm]), temperature 1 (20C) and
temperature 2 (600C). We also know that the volumes are equal in this case (the volume of
the can. First, we can eliminate the volume factor from the equation.
p1/T1 = p2/T2
We just need to rearrange the formula to solve for the new pressure.
p2 = (p1/T1) X T2 = [(1 atm)/20C] X 600C = 30 atm
30. A total of 500 kJ of heat is added to 1 kg of water at 20C.
What is the final temperature
of the water? If it is 100C, how much steam, if any is produced?
Relevant formulas:
Q = mcT (heat added = mass X specific heat X temperature change)
And
Hv = heat of vaporization
( = amount of energy needed to convert 1 kilogram of a substance from a liquid into a
vapor [ ie: water into steam])
We are given the overall amount of energy input (500 kJ), the mass of water (1 kg) and the
change in temperature (100C - 20C = 80C). First we solve for the amount of energy
needed to raise 1 kg of water by 80C:
Q = mcT = 1 kg X 4.2 kJ/kgC X 80C = 336 kJ
To find out how much energy we have left over, we subtract the energy required to raise the
1-kg of water from 20C to 100C (336 kJ) from our total energy input (500 kJ):
Energy remaining = 500 kJ – 336 kJ = 164 kJ
Hv for water is 2260 kJ/kg. To find the amount of water converted into steam, we set the total
energy equal to the heat of vaporization X the mass of water that gets converted to steam.
Energy = Hv X m
Rearranging to solve for the mass of water converted to steam:
m = Energy/Hv = (164 kJ)/(2260 kJ/kg) = 0.0726 kg = 72.6 g of water converted to steam
38. An engine that operates at the maximum efficiency possible takes in 1.0 MJ (megajoules
= 106 joules) of heat at 327C and exhausts waste heat at 127C. How much work does it
perform?
From pg 137 we get the relevant formulas:
Maximum efficiency = (work output)/(energy input) = Eff(max)
And
Eff(max) = 1 – Tcold/Thot
We are given the energy input (1.0 MJ), the Tcold (127C = 400K), and the Thot (327C =
600K). Substituting in and solving for Eff(max):
Eff(max) = 1 – Tcold/Thot = 1 - 400K/600K = 1 - .667 = 0.333
Now, rearranging the first formula to solve for the work output:
Work output = Eff(max) X (energy input) = 0.333 X 1.0 X 106 J = 333,000 J or 333 kJ