Practice Final 2
Problem 1
How many slices of pizza must you eat to walk for 1.0 h at a speed of 5.0 km/h?
Assume (i) energy content of a slice of pizza is 300 Cal, (ii) metabolic power is 380 W,
and (iii) human efficiency is 25%. (Answer: 4.3)
Solution
Walking at 5 km/hr requires about 380 W of metabolic power and with the usual
efficiency at 25%, the usable energy is
e  usable energy  power  time  (380 W)  (3600s)  1370kJ
1370kJ
usable energy 
 5480 kJ
0.25
A typical slice of pizza has an energy content of 300 Cal = 1260 kJ. How much slices "n"
of pizza are required to supply the amount of useable energy is the total energy you eat
is n times 1.26 kJ. Solving for n gives
n  (energy content of pizza)  usable energy
n  (1260 kJ)  5480kJ
5480 kJ
n
 4.3  n
1260 kJ
Problem 2
A weather balloon rises through the atmosphere, its volume expanding from 4.0 m3 to 12
m3 as the temperature drops from 20oC to −10oC. (a) If the initial gas pressure inside the
balloon is 1.0 atm, what is the final pressure? (b) Does the pressure decrease? Explain
your answer using short concise sentences. (Answer: 0.30 atm)
Solution
We are given the initial volume, temperature and pressure of the gas are V1 = 4.0 m3,
T1 = 20oC = 293 K, p1 = 1 atm; the final volume and temperature are V2 = 12 m3, T2 =
−10oC = 263 K. We can rewrite the ideal gas law
pV p V
nR  constant  1 1  2 2
T1
T2
Solving for p2:
T V
4.0 m3 263 K
p2  p1  2  1  (1.0 atm) 

 0.30 atm  p2
T1 V2
12 m3 293 K
Problem 3
A 68 kg man dances with his date for 1.5 h, using a metabolic power of 250 W (assume
efficiency of 25%). All of the metabolic energy that does not go to dancing s converted to
thermal energy in her body. If he couldn't perspire or otherwise cool his body, by how
much would his body temperature rise during this exercise? Assume a specific heat of
cbody = 3400 J/kg∙K. (Answer: 4.39oC)
Solution
As the 68 kg man exercises, the heat produced goes to increasing the thermal energy of
the man (∆Eth = mc∆T). We need to find the temperature increase, given that his
metabolic power efficiency is 0.75 ∙P = ∆Eth/∆t = 0.75∙250 J/s = 188 J/s is equal to the
rate of heat energy increase. Note: the time of exercise ∆t = 90 min = 5400s, and the
man’s mass 68 kg. For the specific heat, we use c = 3400 J/kg∙K. Solving first for the
thermal energy produced by the exercise, the power equation gives us
Eth
P
 Eth  Pt  (188 W)(5400 s)  1.02  106 J
t
We can solve the equation ∆Eth = mc∆T for the change in temperature ∆T:
(1.02  106 J)
T  Eth / mc 
 4.39 K  4.39 C  7.9 F  T
(68 kg)(3400 J/(kg  K))
Problem 4
A Styrofoam box (ksty = 0.01 W/m∙K) used to keep drinks cold at a picnic has total wall
area (including the lid) of 0.80 m2 and wall thickness 2.0 cm.
a. What is the rate of heat flow into the box if the temperature of the outside wall is
30oC?
b. How much ice melts in one day? (Lf = 3.34 ×105 J/kg) (Answer: 12 W, 3.1 kg)
Solution
a. The rate of conduction across a temperature difference is
Q  kA 
(0.01W/(m  K))( 0.8m2 )
Pconduction 
 
(30  0C)
 T 
t  L 
0.020 m
Q
 Pconduction
t
b. The total heat flow Q in one day (86,400 s) is
Q  Pconductiont  12 J/s  86,400s  1.04 106 J
So the quantity of ice melted by this quantity of heat is
Q
1.04 106 J
QL  mLf  m  L 
 3.1 kg  m
Lf 3.34 105 J / kg
Problem 5
A 50-cm-thick layer of oil (ρoil = 900 kg/m3) floats on a 120-cm-thick layer of water. What
is the pressure at the bottom of the water layer? (Answer: 120 kPa)
 12 J/s  12 W 
Solution
The pressure at the bottom of the oil layer is the weight of air
plus oil plus water:
pbottom  patm  oilgd1  water gd2
 (1.013  105 Pa)  (900 kg/m3 )(9.80 m/s2 )(0.50 m)
 (1000 kg/m3 )(9.80 m/s2 )(1.20 m)
 1.2  105 Pa  1.2atm  pbottom
Problem 6
A 200 g air-track glider is attached to a spring. The glider is pushed 10.0 cm against the
spring. then released. A student with a stopwatch finds that 10 oscillations take 12.0 s.
What is the spring constant? (Answer: 5.5 N/m)
Solution
The air-track glider attached to a spring is in simple harmonic motion. Experimentally,
the period of motion is
12.0 s
T
 1.20 s
10 oscillations
Using the equation for the period,
2
T  2
2
 2 
 2 
m
 k  
 (0.200 kg)  5.5 N/m  k
 m
k
 T 
 1.20 s 
Problem 7
Use short concise sentences to answer the following questions using physics principles.
Make sure to explain your reasoning for each answer
a. Why does food cook faster in a pressure cooker than in an open pot of boiling water?
b. A person pours a cup of hot coffee, intending to drink it five minutes later. To keep
the coffee as hot as possible, should she put cream in it now, or wait until just before
she drinks it?
c.
Before giving you an injection, a physician swabs your arm with isopropyl alcohol at
room temperature. Why does this make your arm feel cold? The boiling point of
isopropyl alcohol is 82.4oC.
d. During the Great Mississippi flood of 1993, the levees in St. Louis tended to rupture
first at the bottom. Why?
e. Why do short dogs (like Chihuahuas) walk with quicker strides than do tall dogs (like
Great Danes)?
f. The frequency with which dogs pant is the natural frequency of their respiratory
system. Why do they choose this frequency?
e
E
what you get
 output
what you had to pay Einput
K total  NK avg  Eth  32 kB T
P
T(C) 
pV  nRT
Q  McT
ρ  m/V
Q
Power 
Q kA

T
t
L
Qf,v  ML f,v
p  F/A
T(F)  32
patm  1 atm  101 kPa
V
 v1A1  v 2 A 2
t
T(K)  T(C)  273
kB  1.38  1023 J/K
1 Cal  1 kcal  4.19 kJ
p  F/A
5
9
W E

t
t
E
t
Q
 e AT 4
t
pgauge  p  1 atm
p1  21  v12   gy1  p2  21  v 22   gy 2
FS  kx
TSpring  2
v  f
v
F
m/L
m
k
TPendulum  2
L
g
 v 
fn  nf1  n  
 2L 