Given it's already ice, latent heat is not a factor. So any further cooling goes into lowering the temperature as in dT Cp M = dQ so that dQ = 40*2090*.4 = 3.34400E+04 Joules. ANS.
Note C and K have the same interval in degrees so we can use either in dT.
Answers & Comments
40 degrees Celsius is the same thing as 40 K, so the temperature change is 40 K and the mass is 400g which is 0.40 kg.
The units of the specific heat actually suggest how to do the problem: 2090 J per kilogram per kelvin.
The total energy used, then, is just 2090 * 40 * 0.4 = 334,400 J = 334.4 kJ
Given it's already ice, latent heat is not a factor. So any further cooling goes into lowering the temperature as in dT Cp M = dQ so that dQ = 40*2090*.4 = 3.34400E+04 Joules. ANS.
Note C and K have the same interval in degrees so we can use either in dT.
You have 0.4 kg and you want to change the temperature 40°.
0.4kg * 40°C * 2090J/kg-°C = 33440 J