The calorimeter contains 1 kg of ice at a temperature of -10 degrees. What amount of heat must be transferred

The calorimeter contains 1 kg of ice at a temperature of -10 degrees. What amount of heat must be transferred to it so that ice turns into water at a temperature of 20 degrees?

1) First, the ice will heat up from -10 ° C to 0 ° C. This will require energy:
Q1 = cl * ml * (t2 – t1).
Q1 = 2400 J / (kg * ºС) * 1 kg * (0ºС – (-10 ºС)) = 24000 J.
2) Then the ice will begin to melt, this requires energy: Q2 = Ll * ml.
Q2 = 3.3 * 10 ^ 5 J / kg * 1 kg = 330,000 J.
3) Finally, the resulting water is heated from 0 ° C to 20 ° C. This will require energy:
Q3 = cw * ml * (t2 – t1).
Q3 = 4200 J / (kg * ºC) * 1 kg * (20ºC – 0ºC) = 84000 J.
4) In total, it will go: Q = Q1 + Q2 + Q3 = 24000 J + 330000 J + 84000 J = 438000 J.
Answer: 438,000 J.