Problem#1
An energy-efficient lightbulb, taking in 28.0 W of power, can produce the same level of brightness as a conventional bulb operating at power 100 W. The lifetime of the energy efficient bulb is 10 000 h and its purchase price is $17.0, whereas the conventional bulb has lifetime 750 h and costs $0.420 per bulb. Determine the total savings obtained by using one energy-efficient bulb over its lifetime, as opposed to using conventional bulbs over the same time period. Assume an energy cost of $0.080 0 per kilowatt-hour.Answer:
We use power = energy/time then e
For the 28.0 W bulb:
Energy used = (28.0 W)(1.00 x 104 h) = 280 kWh
Total cost = $17.0 + (280 kWh)( $0.0800/kWh) = $39.4
For the 100 W bulb:
Energy used = (100 W)(1.00 x 104 h) = 1.00 x 103 kWh
Bulb used = (1.00 x 104 h)/(750 h/bulb) = 13.3, the
Total cost = 13.3($0.420) + (1.00 x 103 kWh)( $0.0800/kWh) = $85.6
Savings with energy-efficient bulb = $85.6 kWh – $39.4 kWh = $46.2
Problem#2
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kcal # 4 186 J. Metabolizing one gram of fat can release 9.00 kcal. A student decides to try to lose weight by exercising. She plans to run up and down the stairs in a football stadium as fast as she can and as many times as necessary. Is this in itself a practical way to lose weight? To evaluate the program, suppose she runs up a flight of 80 steps, each 0.150 m high, in 65.0 s. For simplicity, ignore the energy she uses in coming down (which is small). Assume that a typical efficiency for human muscles is 20.0%. This means that when your body converts 100 J from metabolizing fat, 20 J goes into doing mechanical work (here, climbing stairs). The remainder goes into extra internal energy. Assume the student’s mass is 50.0 kg. (a) How many times must she run the flight of stairs to lose one pound of fat? (b) What is her average power output, in watts and in horsepower, as she is running up the stairs?
Answer:
(a) Burning 1 lb of fat releases energy
1lb(454 g/lb)(9 kcal/1g)(4186 J/1kcal) = 1.71 x 107 J
The mechanical energy output is
(1.71 x 107 J)(0.20) = nF∆r cosθ
3.42 x 106 J = nmg∆y cos00
3.42 x 106 J = n(50.0 kg)(9.80 m/s2)(0.150 m)(80 steps) cos00
n = (3.42 x 106 J)/(5.88 x 103 J) = 582
This method is impractical compared to limiting food intake.
(b) Her mechanical power output is
P = W/∆t = (5.88 x 103 J)/(65 s) = 90.5 W = 0.121 hp
Problem#3
For saving energy, bicycling and walking are far more efficient means of transportation than is travel by automobile. For example, when riding at 10.0 mi/h a cyclist uses food energy at a rate of about 400 kcal/h above what he would use if merely sitting still. (In exercise physiology, power is often measured in kcal/h rather than in watts. Here 1 kcal = 1 nutritionist’s Calorie = 4 186 J.) Walking at
3.00 mi/h requires about 220 kcal/h. It is interesting to compare these values with the energy consumption required for travel by car. Gasoline yields about 1.30 x 108 J/gal. Find the fuel economy in equivalent miles per gallon for a person (a) walking, and (b) bicycling.
Answer:
(a) The fuel economy for walking is
(1 h/220 kcal)(3 mi/h)(1 kcal/4186 J)(1.30 x 108 J/1 gal) = 423 mi/gal
(b) For bicycling
(1 h/400 kcal)(10 mi/h)(1 kcal/4186 J)(1.30 x 108 J/1 gal) = 776 mi/gal
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