Given the Following: heat of fusion of ice = 80.0 cal/g; specific heat of ice = 0.480 cal/g; and specific heat of water = 1.00 cal/g. Calculate the amount of heat required to melt 30.0 g of ice at 0 degrees celsius?

1 Answer
Jul 4, 2017

Answer:

We wish to assess the enthalpy change for the physical reaction......

Explanation:

#H_2O(s) +DeltararrH_2O(l)#.......

Note that BOTH the ice and water have a specified temperature of #0# #""^@C#, and thus we assess the heat involved in the phase change, and thus we only need the #"latent heat of fusion of ice"#, #80.0*cal*g^-1#.

And for the given quantity, this #80*cal*g^-1xx30*g=2400*cal#; whatever a #"calorie"# is...........

By the same token, we would report the enthalpy change of the reverse reaction......

#H_2O(l) +DeltararrH_2O(s)#, #DeltaH^@=-2400*cal#. Do you agree?