Question

How much energy is required to change a 50-g ice cube from ice at −35°C to steam at 125°C? (in MJ)

Answer 1

converting ice from -35 degrees celcius to 0 degrees celcius
using the formula Q=mc * delta T
Q=0.05kg (2220) *(0-(-35)
Q=3885J
converting ice from 0 degrees celcius to water at 0 degrees celcius

Q=ml
Q=(0.05) * (333000) =16650J

converting water at 0 degrees celsius to water at 100 degrees celcius

Q=mc(t2-t1)
Q=0.05 * 4200 * (100-0)
Q=21000J

converting water at 100 degrees celcius to steam at 125 degrees celcius
Q=mc(t2-t1)
Q=0.05 * 2256000 * (125-100)
Q=2820000J = 2820 kJ

Qtotal=2820000J+21000J +16650J +3885J =2861535 joules