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Ok, can someone please tell me what the specific heat capacity is for water...
I know I can look up in the internet but my net is messed up at the moment...
Anyways can you give me it in kj/kg AND j/g ? Thanks a lot

2006-10-25 08:57:55 · 9 answers · asked by A 2 in Science & Mathematics Chemistry

CHESSLARUS you gave this as an answer:
Cp = 4.1855 J / g K (at 15 °C)
Cp = 4.1855 kJ / kg K (at 15 °C)
But why is the specific heat capacity the same for in both J/g AND kJ/Kg ?
Please some answer me.

2006-10-25 09:13:41 · update #1

9 answers

4.184 J/g...we just started this in chem today too

2006-10-25 09:11:08 · answer #1 · answered by Anonymous · 0 0

For water:

Cp = 4.1855 J / g K (at 15 °C)
Cp = 4.1855 kJ / kg K (at 15 °C)

That´s It!
Good luck!

2006-10-25 09:07:37 · answer #2 · answered by CHESSLARUS 7 · 0 0

THat answer is Cp = .239 J / g K
This means that you need 4.1855 J of energy to rise 1 °C of One gram of water .
Other one is Cp = 4.1855 kJ / kg : meaning you need 4.1855 KJ to rise 1 °C of One Kilogram of water.

This answer is right because I am a chemistry teacher.

2006-10-25 10:22:36 · answer #3 · answered by malutty92 2 · 0 0

The values are the same in J/g and kJ/kg because you are multiplying by 1,000 and dividing by 1,000 to convert - so the numerical value stays the same.

2006-10-29 05:41:39 · answer #4 · answered by Gervald F 7 · 0 0

4.18 J/g C
If you are doing a problem with kJ personally i would convert the numbers in your problem to J,,,it would be a lot easier

2006-10-25 10:31:32 · answer #5 · answered by need4speed 2 · 0 0

the specific heat capacity of water is the following
4184 J/(kg·K) at a liquid

2006-10-25 09:12:05 · answer #6 · answered by gordon_benbow 4 · 0 0

Make a simple compare about your blood count. Those who have more will less vulnerable to dry cold.

2016-03-28 07:27:28 · answer #7 · answered by Anonymous · 0 0

4.2 J/Kg
0.0042 J/g
1 Cal/Kg

2006-10-25 09:31:31 · answer #8 · answered by T-wad 2 · 0 0

4.186J/g. You convert.

2006-10-25 09:07:23 · answer #9 · answered by Anonymous · 0 0

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