Hot Steel; Cold Ice: A 36.0 g block of ice sits at −50.0 °C under 1.00 atm constant pressure in a perfectly insulated container. A 225 g mass of stainless steel at 125 °C is dropped onto the ice, and the system is closed. When the system reaches thermal equilibrium, ice remains. What is the value of q_steel? Enter your answer in Celsius using no decimal places

Question

Note that:
delta H fus (H2o) = 333 j/g
delta H vap (H2o) = 2260 J/g
Cp (steel) = 0.500 J g^-1°C^-1
S(H2o(s)) = 2.11 J g^-1 °C^-1
S (H2o(f)) =4.18J g^-1 °C^-1
S (H2o(g)) = 2.00 J g^-1 °C^-1

Answer 1

Heat is in Joules, not degrees Celsius.
If some ice is still there then it is at 0 deg C
Anyway the steel is cooled from 125 C to 0 C
heat out of steel = 225 g (125 - 0) deg C * 0.500 Joules / g deg C

Answer 2

so would it be 14062?

Answer 3

That is what I get.

Answer 4

Remember it is heat out so you should call it negative.