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View Full Version : Newton's Law of Cooling. little question about 'Specific heat' of Mercury


matthewgill
Apr15-09, 09:17 AM
Newton's Law of Cooling

Time Required to Cool (or heat) an Object [Also known as Newton's Law of Cooling]

t=mCp(Ts-Tf)/Q

Where;

Q is the heat added (or being removed) from the object in watts
m is the mass (weight) of the object in Kg
Cp is the Specific heat of the object material in J/Kg C
t is the time required to cool down (or heat up) the object in seconds
Ts is the starting temperature in C
Tf is the final temperature in C

Here is what i am entering:

Q - 1700 W
m - 500 KG
Cp - ??????
Ts - 1 degC
Tf - 67 degC

But what should i enter for the 'Specific heat of the object material in J/Kg C'?
The Object is Mercury.

Thank you

Bob S
Apr15-09, 01:18 PM
Cp=0.139 joules per gram per degree C.