1. You stop your bicycle by squeezing on the brake
handles, which causes 4 rubber pads to rub against
the metal rims of the wheels. Suppose you were
going 10mph and stopped the bicycle that way
in 10 meters. If you and your bicycle have a mass
of 80kg, how much force in the direction of
the rim's motion did each brake pad
exert on a wheel? Ignore any effects not due to
the brakes which might cause the bike to slow down.
1mi=1609 meters
A. (1/8) 10 x 80/102 Newtons
B. (½) 102 /8000 Newtons
C. (1/8) 800(1609/3600)2 Newtons
D. (1/8) 800(3600/1609)2 Newtons

2. Initial energy = work done by friction to stop:
Answer: C
Initial energy = work done by friction to stop:
(1/2)Mv2 =4Fd
Solve for F and insert M=80kg, v =10mph =10(1609/3600)m/s,
d=10m. F=
(1/8) 800(1609/3600)2 Newtons =19.9 N

3. Continuing the previous problem. Suppose that
each metal rim of the bicycle has mass 1kg, that
the wheels are made of titanium
and that any thermal energy added to the rims
during the stop does not have time to leak out
into the environment by the time you are completely
stopped. At the moment when you stop, how
much has the temperature of the rims increased?
Specific heat of titanium = 523 joules/kgoC.
A. 19.9/523 centigrade degrees
B. 199/523 centigrade degrees
C /5230 centigrade degrees
D. 523/ centigrade degrees

4. Answer: B
(Change in T)x(mass)x(specific heat)
= increase in thermal energy =Fd
Gives temperature change of
19.9 newtons x 10m/((523 joules/kg degree)x1kg)=
194/523 centigrade degrees=.38 centigrade degrees

5. You have a solar cooker like the one in Figure 6.16 of the
book. The R value of the glass surrounding the food is 1 ft2 Fhr/btu. It lets all the solar radiation pass through but, like a greenhouse, none of the infrared radiation from the hot food gets out. 50% of the solar radiation of 600 watts/m2 which is incident on the mirrors
hits the black can containing the food. The black can completely absorbs the radiation which hits it. The mirrors have area .25 m2.
The outside temperature is 50F The surface area of the container
for the food is 1 ft2 .
1 btu=1055 joules. T(Fahrenheit)=T(C)x(9/5)+32.
What is the rate at which solar electromagnetic energy comes into the food container (in watts)?
A. 300watts B. 75 watts C watts D watts

6. Answer: B
600watts/m2 x(1/4 m2 )x(1/2)

7. You have a solar cooker like the one in Figure 6. 16 of the book
You have a solar cooker like the one in Figure 6.16 of the book. The R value of the glass surrounding the food is 1 ft2 Fhr/btu. It lets all the solar radiation pass through but, like a greenhouse, none of the infrared radiation from the hot food gets out. 50% of the solar radiation of 600 watts/m2 which is incident on the mirrors hits the black can containing the food. The black can completely absorbs the radiation which hits it. The mirrors have area .25 m2.The outside temperature is 50F The surface area of the container for the food is 1 ft2 .
1 btu=1055 joules. T(Fahrenheit)=T(C)x(9/5)+32
Assuming that the 75 watt input is balanced by heat conduction out of the food container, what is the temperature of the food in the cooker (in Fahrenheit) ?
A (3600/1055)-50 F
B (1055/3600)+50 F
C (3600/1055)+50 F
D (3600/1055)-50 F

8. Answer: C
(T-50)F x1ft2 /(1ft2hr/btu)=
75joules/s x (3600 s/hr)/(1055 joules/btu)
Solve for T giving T=122 F

9. A nuclear 150 megawatt (useful output)
electrical generating plant has a thermal
efficiency of 25%
How much thermal energy is coming into the plant (in meggawatts)?
A megawatts
B megawatts
C megawatts
D megawatts

10. Answer: B
Thermal input x 0.25 = 150 Mwatts
Thermal input =600 MWatts

11. A nuclear 150 megawatt (useful output)
electrical generating plant has a thermal efficiency of 25%
If the cooling water used for the nuclear plant
were required to only raise the temperature by 2o C after it absorbs the waste heat, how much water per hour would be required to cool the nuclear plant (in cubic meters/hr) Heat capacity of water = 4186 kJ/m3oC. A.
450(3600)/(2(4.186) )m3/hr
B. 150(3600)/(2(4.186) )m3/hr C.
D. 600(3600)/(2(4.186) )m3/hr
450(2(4.186) )/3600 m3/hr
450(3600)/(2(4.186) )m3/hr
B. 450(3600)/(2(4.186) )m3/hr C. D.

12. Answer: A
Waste thermal energy/s = = 450 Mega watts
=2o C x (volume of water per hour) x 4186 kJ/hr
/3600 s/hr So
Volume of water per hour =
450x106 (3600/(4186 x1000x 2))=
450x3600/(2x4.186) m3 /hr =193,502 m3 /hr

13. A nuclear 150 megawatt (useful output)
electrical generating plant has a thermal efficiency of 25%.
If the temperature of the cooling water is 15o C, what is the minimum temperature that the steam coming into the generator from the hot side
can be?
A. 20o C
B. 100o C
C. 111 o C
D. 60o C

14. Answer: C
(1-Tc/Th )=0.25
Tc/Th =0.75
Th =(4/3)x(15+273)=384 o K= 111 o K C

15. In class, I boiled some water in a demonstration.
If I started with 1kg of water and it all
boiled away in 45 minutes, what was the
rate in watts at which the flame was putting
thermal energy into the water?
Latent heat of vaporization of water =540kJ/kg
A /45
B x45/60
C /(45x60)
D /(45x60)

16. Answer: D
540 kJ/kg x 1 kg =
(output in kwatts) x (45 min x 60 s/min)
So output in watts =
540000/(45x60)= 142,422 watts =142 kwatts