How much will the internal energy of 10 kg of ice taken at a temperature of 0 ° C increase, as a result of its transformation

How much will the internal energy of 10 kg of ice taken at a temperature of 0 ° C increase, as a result of its transformation into steam, at a temperature of 100 ° C?

m = 10 kg.
q = 3.4 * 10 ^ 5 J / kg.
C = 4200 J / kg * ° C.
r = 2.6 * 10 ^ 6 J / kg.
t1 = 0 ° C.
t2 = 100 ° C.
ΔU -?
In order to obtain steam from ice at a temperature of t2 = 100 ° C, it is first necessary to convert the ice at the melting point into a liquid state, heat the resulting water from t1 to the boiling point t2, and convert it to steam at the boiling point.
All the reported amount of heat Q is used to increase the internal energy: Q = ΔU.
ΔU = q * m + C * m * (t2 – t1) + r * m.
ΔU = 3.4 * 10 ^ 5 J / kg * 10 kg + 4200 J / kg * ° C * 10 kg * (100 ° C – 0 ° C) + 2.6 * 10 ^ 6 J / kg * 10 kg = 33600000 J.
Answer: the internal energy of steam will be ΔU = 33,600,000 J more than ice.