| Alternative Energy Demystified, 2nd edition |
| Stan Gibilisco |
| Explanations for Quiz Answers in Chapter 3 |
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1. To convert a temperature reading from degrees Fahrenheit (ºF) to its equivalent in degrees Celsius (ºC), we use the formula C = (5/9) (F - 32) where F represents the Fahrenheit value and C represents the Celsius value. In this case F = -10, so C = (5/9) (-10 - 32) rounded off to the nearest degree Celsius. The correct choice is B. |
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2. To convert a temperature reading from degrees Celsius (ºC) to its equivalent in degrees Fahrenheit (ºF), we use the formula F = (9/5) C + 32 where C represents the Celsius value and F represents the Fahrenheit value. In this case C = +25, so F = (9/5) (+25) + 32 rounded off to the nearest degree Fahrenheit. The correct choice is D. |
| 3. Let's convert from kelvins (K) to degrees
Celsius (ºC), and then to degrees Fahrenheit (ºF). To get the Celsius
value from the Kelvin value, we must subtract 273.15, because kelvins
always exceed degrees Celsius by 273.15. The kelvin value is 233 K, so the
Celsius value equals 233 - 273.15, or -40.15ºC. (Let's leave in the
fractional part of the degree, so we won't have problems with rounding
errors in our arithmetic. We'll round off to the nearest degree Fahrenheit
at the end of the process.) Now we use the formula
F = (9/5) C + 32 In this case C = -40.15, so
F = (9/5) (-40.15) + 32 which equals -40ºF, rounded off to the nearest degree Fahrenheit. The correct choice is A. |
| 4. In scientific applications, we define standard temperature as 0ºC. That's 273.15 K, which rounds off to 273 K. The answer is C. |
| 5. If we symbolize the heat of vaporization for a particular substance
(in calories per gram) as hv, the heat energy added
or given up by a sample of that substance (in calories) as h, and
the mass of the sample (in grams) as m, then
hv = h/m In this situation, we have hv = 126 cal/g and m = 100 g. When we input these values into the above formula, we get 126 = h/100 We can multiply both sides of this equation by 100 and then switch it around left-to-right, solving it as h = 126 x 100 The correct choice is B. |
| 6. This situation represents precisely the reverse process from the one described in Question 5. The sample will lose the same amount of heat energy as it gained before. The answer is B. |
| 7. Air-exchange heat pumps work best when the outdoor air temperature exceeds approximately +4ºC or 277 K. The correct choice is A. |
| 8. If someone claims that a heat pump works with "250 percent efficiency," she most likely refers to the coefficient of performance (COP). No machine can operate with greater than 100 percent efficiency; you never get something for nothing in physics. However, a heat pump can move more energy from one place to another than it takes to do the job. A COP figure of 2.5 (or 250 percent ) lies within the realm of practicality. In fact, some heat pumps have COP specifications higher than that. Under ideal conditions, a good heat pump can move three or four times as much thermal energy from the outdoors to the indoors as it demands from the electric utility in the same units (such as joules). The correct choice is D. |
| 9. An electric resistance heating element gives off radiant energy almost entirely in the form of infrared (IR). The correct choice is C. |
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10. In order to find the dissipated power PBtu/h in British thermal units per hour for a resistance heating element supplied with E volts RMS and having a resistance of R ohms, we use the formula PBtu/h = 3.41 E 2 / R When we input the known values E = 117 and R = 11.7, we get PBtu/h
= 3.41 x 1172 / 11.7 The correct choice is B. |