Calculate the amount of heat (in J) required to heat 2.00 grams of ice from –12.0 °C to 88.0 °C.

The heat of fusion of water is 6.02 kJ/mol, the specific heat capacity of ice is 2.09 J/g ▪°C and the specific heat capacity of water is 4.184 J/g ▪°C.

Can someone help me understand the process of answering these questions... and a way to remember the process that isn't confusing

Question

Calculate the amount of heat (in J) required to heat 2.00 grams of ice from –12.0 °C to 88.0 °C.

The heat of fusion of water is 6.02 kJ/mol, the specific heat capacity of ice is 2.09 J/g ▪°C and the specific heat capacity of water is 4.184 J/g ▪°C. 

Can someone help me understand the process of answering these questions... and a way to remember the process that isn't confusing

anonymous · Answer

Ok so now you have ice initially at -12 degree and you know ice melts at 0. Do you know the equation Q=mc (delta T)?

anonymous · Answer

Q = 2 * 2.09 * 100 = 418 J

anonymous · Answer

lol thats wrong ice melts at 0 degrees.

anonymous · Answer

Yes I do... Do i plug in 0 where c would be?

anonymous · Answer

Ok so first find heat required to change ice from -12 to 0. Then find Heat required for ice to melt(m*L). Then find heat required to change water from 0 to 88