Q = mHf
where Q is the transfer of thermal energy, m is the mass of the ice, and Hf is the specific heat of fusion for melting/freezing.
Hf = 333kJ/kg
m = 1kg
Q = mHf
Q = 1kg*333kJ/kg
Q = 333kJ
1 Joule = 0.239cal
Therefore, multiply 333kJ *0.239kcal/kJ = 79.5kcal. Significant digits forces it to 80kcal.
1 kg block of ice at 0 degrees celsius placed in cooler. How much heat will the ice remove as it melts to water at 0 degree celsius. I know the answer is 80kcal but don't know the math that backs up the answer-please help!
2 answers
it is right, but why not work it in cal from the start. The heat of fusion for ice is 80kcal/kg ?
q=mLf=1kg*80kcal/kg
q=mLf=1kg*80kcal/kg
1 answer