Question

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?

Answer 1

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?