Temperature
Practice
practice problem 1
- Derive the equations to convert temperatures in…
- degree Fahrenheit to degree Celsius
- degree Celsius to degree Fahrenheit
- degree Fahrenheit to kelvin
- kelvin to degree Fahrenheit
- At what temperature do the…
- Celsius and Fahrenheit scales coincide?
- Kelvin and Fahrenheit scales coincide?
- Complete the table below.
kelvin | Celsius | Fahrenheit | |
---|---|---|---|
absolute zero | +0,000.000 | ||
100 K | +0,100.000 | ||
[°C] = [°F] | |||
0 °F | +0,000.000 | ||
n.f.p. H2O | +0,273.150 | +0,000.000 | +0,032.000 |
room temp. 1 | +0,020.000 | ||
room temp. 2 | +0,300.000 | ||
body temp. | +0,037.000 | ||
100 °F | +0,100.000 | ||
n.b.p. H2O | +0,373.150 | +0,100.000 | +0,212.000 |
[K] = [°F] | |||
1,000 °F | +1,000.000 | ||
1,000 K | +1,000.000 | ||
1,000 °C | +1,000.000 |
solution
Deriving temperature scale conversions is like determining a linear equation from its slope and intercept.
Between the freezing and boiling points of water, there are 100 − 0 = 100 Celsius degrees for every 212 − 32 = 180 Fahrenheit degrees. This is the same as multiplying the interval by 59. But before we do that we we need to align the the lower fixed point of the Fahrenheit to the zero of the Celsius scale by subtracting 32. Something like this…
y = mx + b [°C] = 100 − 0 ([°F] − 32) 212 − 32 This time we multiply by 95 (the reciprocal). Since the Fahrenheit degree is smaller than the Celsius degree there are more of them in any interval. And since the lower fixed point of the Celsius scale is already zero we don't need to add 32 until after multiplying. The reverse conversion then looks like this…
y = mx + b [°F] = 212 − 32 [°C] + 32 100 − 0 There is no interesting way to convert degrees Fahrenheit to kelvin. Just convert it to degrees Celsius first and then add 273.15.
[K] = [°C] + 273.15 The reverse conversion is similarly boring. Convert kelvin to degree Celsius by subtracting 273.15, then convert that to degree Fahrenheit.
[°F] = 95[°C] + 32
Asking when two temperature scales coincide is like asking when a linear equation is equal to itself. (Note that I did not ask when the Celsius and kelvin scales coincide. That problem has no solution.)
For the Celsius and Fahrenheit scales we get…
[°C] = 59([°F] − 32) x = 59(x − 32) 9x = 5x − 5(32) or…
[°F] = x = 95x + 32 5x = 9x + 9(32) For the kelvin and Fahrenheit scales we get…
[K] = 59([°F] − 32) + 273.15 x = 59(x − 32) + 273.15 9x = 5x − 5(32) + 9(273.15) 4x = 9(273.15) − 5(32) or…
[°F] = 95([K] − 273.15) + 32 x = 95(x − 273.15) + 32 5x = 9x − 9(273.15) + 5(32) −4x = 5(32) − 9(273.15)
Finally, complete the table below.
kelvin | Celsius | Fahrenheit | |
---|---|---|---|
absolute zero | +0,000.000 | +0,−273.150 | +0,−459.670 |
100 K | +0,100.000 | +0,−173.150 | +0,−279.670 |
[°C] = [°F] | +0,233.150 | +0,0−40.000 | +0,0−40.000 |
0 °F | +0,255.372 | +0,0−17.778 | +0,000.000 |
n.f.p. H2O | +0,273.150 | +0,000.000 | +0,032.000 |
room temp. 1 | +0,293.150 | +0,020.000 | +0,068.000 |
room temp. 2 | +0,300.000 | +0,026.850 | +0,080.330 |
body temp. | +0,310.150 | +0,037.000 | +0,098.600 |
100 °F | +0,310.928 | +0,037.778 | +0,100.000 |
n.b.p. H2O | +0,373.150 | +0,100.000 | +0,212.000 |
[K] = [°F] | +0,574.588 | +0,301.438 | +0,574.588 |
1,000 °F | +0,810.928 | +0,537.778 | +1,000.000 |
1,000 K | +1,000.000 | +0,726.850 | +1,340.330 |
1,000 °C | +1,273.150 | +1,000.000 | +1,832.000 |
practice problem 2
solution
Answer it.
practice problem 3
solution
Answer it.
practice problem 4
solution
Answer it.