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

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

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. 19.9/5230 centigrade degrees D. 523/199 centigrade degrees

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

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. 316500 watts D. 0.284 watts

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

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. 75(3600/1055)-50 F B. 75(1055/3600)+50 F C. 75(3600/1055)+50 F D. 75(3600/1055)-50 F

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

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. 37.5 megawatts B. 600 megawatts C 450 megawatts D. 150 megawatts

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

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.

Answer: A Waste thermal energy/s = 600-150= 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

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

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

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. 540/45 B. 540x45/60 C. 540/(45x60) D. 540000/(45x60)

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