- #1

- 26

- 0

1lb = 1/2.2046 kg

1 F = 255.928 K

then 1 BTU/lbF should be 1055/(1.2046)*255.928. But I get 9 plugging this in

the value is suppose to be 4186 J/kgK. this is the specific heat of water

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- Thread starter jamesfirst
- Start date

- #1

- 26

- 0

1lb = 1/2.2046 kg

1 F = 255.928 K

then 1 BTU/lbF should be 1055/(1.2046)*255.928. But I get 9 plugging this in

the value is suppose to be 4186 J/kgK. this is the specific heat of water

- #2

SteamKing

Staff Emeritus

Science Advisor

Homework Helper

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You've made a goofy mistake here. You're confusing a 1 degree F temperature difference with a temperature of 1° F.

1lb = 1/2.2046 kg

1 F = 255.928 K

then 1 BTU/lbF should be 1055/(1.2046)*255.928. But I get 9 plugging this in

the value is suppose to be 4186 J/kgK. this is the specific heat of water

On the Fahrenheit scale, there are 180 degrees separating the F.P. of water and its B.P. On the Kelvin scale, there are 100 degrees between the same two temperatures.

Thus, a temperature difference of 1°F = a temperature difference of 100 / 180 = 5/9 °K.

Try again to make your conversion.

- #3

- 26

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hey it works!

thanks. ive been confusing it with the temperature Fahrenheit the whole time ...

thanks. ive been confusing it with the temperature Fahrenheit the whole time ...

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