Water's heat of fusion is 80. cal/g , its specific heat is 1.0calg⋅∘c, and its heat of vaporization is 540 cal/g . a canister is filled with 330 g of ice and 100. g of liquid water, both at 0 ∘c . the canister is placed in an oven until all the h2o has boiled off and the canister is empty. how much energy in calories was absorbed?

Refer to the diagram shown below.

State 1:
The heat of fusion is required for the ice to become water.
The amount of heat required is
Q₁ = (330 g)*(80 cal/g) = 26,400 cal

State 2:
Sensible heat is required to raise water at 0 C to the boiling point of 100 C.
The heat required is
Q₂ = (430g)*(1 cal/(g-C))*(100 C) = 43,000 cal

State 3:
The heat of vaporization is required to boil off the water.
The heat required is
(430 g)*(540 cal/g) = 232,200 cal

Total heat absorbed is
Q₁+Q₂+Q₃ = 301,600

301,600 cal  or 301.6 kcal

It’s c it stays the same. and the atomic number goes up by 1. a neutron has changed into a proton plus an electron. the electron is emitted as a beta particle.