Mixing
Difference between Celsius and Fahrenheit leads to unexpected observations.
$begingroup$
Say city A has a temperature of 54 F(12.2 C) and city B has a temperature of 44 F(6.7 C).
Their difference is 10 F which is about -12.2 C. I mean what!? This means that 44 F(6.7 C) > 54 F(12.2 C) or 12.2 C - 6.7 C = -12.2C. How is this possible? Is this question silly??
unit-of-measure
asked Jan 25 at 16:54
Vasu090Vasu090
224
$endgroup$
add a comment |
$begingroup$
Say city A has a temperature of 54 F(12.2 C) and city B has a temperature of 44 F(6.7 C).
Their difference is 10 F which is about -12.2 C. I mean what!? This means that 44 F(6.7 C) > 54 F(12.2 C) or 12.2 C - 6.7 C = -12.2C. How is this possible? Is this question silly??
unit-of-measure
asked Jan 25 at 16:54
Vasu090Vasu090
224
$endgroup$
$begingroup$
In the Kelvin scale, A has 285.2K and B has 279.7K. Do you see why their difference is not 261.2K (which would correspond to 10°F) and also not 278.5K (which would correspond to 5.5°C) or 267.5K (which woul dcorrespond to -5.5°C), but simply 5.5K?
$endgroup$
– Hagen von Eitzen
Jan 25 at 17:01
$begingroup$
Which one is greater: 1 meter above the floor, or 1 foot above the table?
$endgroup$
– Alexey
Jan 25 at 17:19
add a comment |
$begingroup$
Say city A has a temperature of 54 F(12.2 C) and city B has a temperature of 44 F(6.7 C).
Their difference is 10 F which is about -12.2 C. I mean what!? This means that 44 F(6.7 C) > 54 F(12.2 C) or 12.2 C - 6.7 C = -12.2C. How is this possible? Is this question silly??
unit-of-measure
asked Jan 25 at 16:54
Vasu090Vasu090
224
$endgroup$
Say city A has a temperature of 54 F(12.2 C) and city B has a temperature of 44 F(6.7 C).
Their difference is 10 F which is about -12.2 C. I mean what!? This means that 44 F(6.7 C) > 54 F(12.2 C) or 12.2 C - 6.7 C = -12.2C. How is this possible? Is this question silly??
unit-of-measure
unit-of-measure
asked Jan 25 at 16:54
Vasu090Vasu090
224
asked Jan 25 at 16:54
Vasu090Vasu090
224
asked Jan 25 at 16:54
Vasu090Vasu090
224
asked Jan 25 at 16:54
Vasu090Vasu090
224
asked Jan 25 at 16:54
Vasu090Vasu090
224
224
$begingroup$
In the Kelvin scale, A has 285.2K and B has 279.7K. Do you see why their difference is not 261.2K (which would correspond to 10°F) and also not 278.5K (which would correspond to 5.5°C) or 267.5K (which woul dcorrespond to -5.5°C), but simply 5.5K?
$endgroup$
– Hagen von Eitzen
Jan 25 at 17:01
$begingroup$
Which one is greater: 1 meter above the floor, or 1 foot above the table?
$endgroup$
– Alexey
Jan 25 at 17:19
add a comment |
$begingroup$
In the Kelvin scale, A has 285.2K and B has 279.7K. Do you see why their difference is not 261.2K (which would correspond to 10°F) and also not 278.5K (which would correspond to 5.5°C) or 267.5K (which woul dcorrespond to -5.5°C), but simply 5.5K?
$endgroup$
– Hagen von Eitzen
Jan 25 at 17:01
$begingroup$
Which one is greater: 1 meter above the floor, or 1 foot above the table?
$endgroup$
– Alexey
Jan 25 at 17:19
$begingroup$
In the Kelvin scale, A has 285.2K and B has 279.7K. Do you see why their difference is not 261.2K (which would correspond to 10°F) and also not 278.5K (which would correspond to 5.5°C) or 267.5K (which woul dcorrespond to -5.5°C), but simply 5.5K?
$endgroup$
– Hagen von Eitzen
Jan 25 at 17:01
$begingroup$
In the Kelvin scale, A has 285.2K and B has 279.7K. Do you see why their difference is not 261.2K (which would correspond to 10°F) and also not 278.5K (which would correspond to 5.5°C) or 267.5K (which woul dcorrespond to -5.5°C), but simply 5.5K?
$endgroup$
– Hagen von Eitzen
Jan 25 at 17:01
$begingroup$
Which one is greater: 1 meter above the floor, or 1 foot above the table?
$endgroup$
– Alexey
Jan 25 at 17:19
$begingroup$
Which one is greater: 1 meter above the floor, or 1 foot above the table?
$endgroup$
– Alexey
Jan 25 at 17:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If two scales have non-coincident zeros, the quantity differences will have transformation law that is different from quantity itself:
$$
y =ax+b,qquad text{but}qquad Delta y = y_2-y_1 = (ax_2+b)-(ax_1+b)=a(x_2-x_1)=aDelta x
$$
For Fahrenheit and Celsius:
$$
F = 1.8C+32,qquad text{but}qquadDelta F = 1.8Delta C
$$
so difference in 10°F is 5.55°C.
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
You're evaluating a difference of temperatures, not a temperature; since there are $9$ degrees Fahrenheit to $5$ degrees Celsius, the difference in temperature of the two towns, in degrees Celsius, is
$$
frac{5}{9}cdot 10approx 5.56
$$
The temperatures yesterday here were: maximum $5$℃, minimum $-3$℃. They correspond to $41$℉/$26.6$℉.
The difference in degrees Celsius is $8$, which corresponds to
$$
frac{9}{5}cdot 8=14.4
$$
degrees Fahrenheit, and this agrees with the given data. The fact that a temperature of $14.4$℉ is about $-9.78$℃ is of no concern.
answered Jan 25 at 17:45
egregegreg
184k1486205
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If two scales have non-coincident zeros, the quantity differences will have transformation law that is different from quantity itself:
$$
y =ax+b,qquad text{but}qquad Delta y = y_2-y_1 = (ax_2+b)-(ax_1+b)=a(x_2-x_1)=aDelta x
$$
For Fahrenheit and Celsius:
$$
F = 1.8C+32,qquad text{but}qquadDelta F = 1.8Delta C
$$
so difference in 10°F is 5.55°C.
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
If two scales have non-coincident zeros, the quantity differences will have transformation law that is different from quantity itself:
$$
y =ax+b,qquad text{but}qquad Delta y = y_2-y_1 = (ax_2+b)-(ax_1+b)=a(x_2-x_1)=aDelta x
$$
For Fahrenheit and Celsius:
$$
F = 1.8C+32,qquad text{but}qquadDelta F = 1.8Delta C
$$
so difference in 10°F is 5.55°C.
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
$endgroup$
add a comment |
$begingroup$
If two scales have non-coincident zeros, the quantity differences will have transformation law that is different from quantity itself:
$$
y =ax+b,qquad text{but}qquad Delta y = y_2-y_1 = (ax_2+b)-(ax_1+b)=a(x_2-x_1)=aDelta x
$$
For Fahrenheit and Celsius:
$$
F = 1.8C+32,qquad text{but}qquadDelta F = 1.8Delta C
$$
so difference in 10°F is 5.55°C.
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
$endgroup$
If two scales have non-coincident zeros, the quantity differences will have transformation law that is different from quantity itself:
$$
y =ax+b,qquad text{but}qquad Delta y = y_2-y_1 = (ax_2+b)-(ax_1+b)=a(x_2-x_1)=aDelta x
$$
For Fahrenheit and Celsius:
$$
F = 1.8C+32,qquad text{but}qquadDelta F = 1.8Delta C
$$
so difference in 10°F is 5.55°C.
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
answered Jan 25 at 17:09
Vasily MitchVasily Mitch
2,6791312
2,6791312
add a comment |
add a comment |
$begingroup$
You're evaluating a difference of temperatures, not a temperature; since there are $9$ degrees Fahrenheit to $5$ degrees Celsius, the difference in temperature of the two towns, in degrees Celsius, is
$$
frac{5}{9}cdot 10approx 5.56
$$
The temperatures yesterday here were: maximum $5$℃, minimum $-3$℃. They correspond to $41$℉/$26.6$℉.
The difference in degrees Celsius is $8$, which corresponds to
$$
frac{9}{5}cdot 8=14.4
$$
degrees Fahrenheit, and this agrees with the given data. The fact that a temperature of $14.4$℉ is about $-9.78$℃ is of no concern.
answered Jan 25 at 17:45
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You're evaluating a difference of temperatures, not a temperature; since there are $9$ degrees Fahrenheit to $5$ degrees Celsius, the difference in temperature of the two towns, in degrees Celsius, is
$$
frac{5}{9}cdot 10approx 5.56
$$
The temperatures yesterday here were: maximum $5$℃, minimum $-3$℃. They correspond to $41$℉/$26.6$℉.
The difference in degrees Celsius is $8$, which corresponds to
$$
frac{9}{5}cdot 8=14.4
$$
degrees Fahrenheit, and this agrees with the given data. The fact that a temperature of $14.4$℉ is about $-9.78$℃ is of no concern.
answered Jan 25 at 17:45
egregegreg
184k1486205
$endgroup$
add a comment |
$begingroup$
You're evaluating a difference of temperatures, not a temperature; since there are $9$ degrees Fahrenheit to $5$ degrees Celsius, the difference in temperature of the two towns, in degrees Celsius, is
$$
frac{5}{9}cdot 10approx 5.56
$$
The temperatures yesterday here were: maximum $5$℃, minimum $-3$℃. They correspond to $41$℉/$26.6$℉.
The difference in degrees Celsius is $8$, which corresponds to
$$
frac{9}{5}cdot 8=14.4
$$
degrees Fahrenheit, and this agrees with the given data. The fact that a temperature of $14.4$℉ is about $-9.78$℃ is of no concern.
answered Jan 25 at 17:45
egregegreg
184k1486205
$endgroup$
You're evaluating a difference of temperatures, not a temperature; since there are $9$ degrees Fahrenheit to $5$ degrees Celsius, the difference in temperature of the two towns, in degrees Celsius, is
$$
frac{5}{9}cdot 10approx 5.56
$$
The temperatures yesterday here were: maximum $5$℃, minimum $-3$℃. They correspond to $41$℉/$26.6$℉.
The difference in degrees Celsius is $8$, which corresponds to
$$
frac{9}{5}cdot 8=14.4
$$
degrees Fahrenheit, and this agrees with the given data. The fact that a temperature of $14.4$℉ is about $-9.78$℃ is of no concern.
answered Jan 25 at 17:45
egregegreg
184k1486205
answered Jan 25 at 17:45
egregegreg
184k1486205
answered Jan 25 at 17:45
egregegreg
184k1486205
answered Jan 25 at 17:45
egregegreg
184k1486205
184k1486205
add a comment |
add a comment |
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$begingroup$
In the Kelvin scale, A has 285.2K and B has 279.7K. Do you see why their difference is not 261.2K (which would correspond to 10°F) and also not 278.5K (which would correspond to 5.5°C) or 267.5K (which woul dcorrespond to -5.5°C), but simply 5.5K?
$endgroup$
– Hagen von Eitzen
Jan 25 at 17:01
$begingroup$
Which one is greater: 1 meter above the floor, or 1 foot above the table?
$endgroup$
– Alexey
Jan 25 at 17:19