How much energy is required to change a 40 g ice cube from ice at −14◦c to steam at 118◦c? the specific heat of ice is 2090 j/kg · ◦ c, the specific heat of water is 4186 j/kg · ◦ c, the specific heat of stream is 2010 j/kg · ◦ c, the heat of fusion is 3.33 × 105 j/kg, and the heat of vaporization is 2.26 × 106 j/kg. answer in units of j?

From -44°C to 0°C: Q1 = mCΔT = 40*(2090/1000)*44 = 3678.4 J
Melting at 0°C: Q2 = mHf = 40*(3.33*10^5/1000) =13320 J
Heating liquid water from 0°C to 100°C: Q3 = mCΔT = 40*(4186/1000)*100 = 16744 J
Vaporizing water to steam at 100°C: Q4 = mHvap = 40*(2.26*10^6/100) = 90400 J
Heating steam from 100°C to 118°C: Q5 = mCΔT = 40*(2010/1000)*18 = 1447.2 J

Energy required, Q = Q1+Q2+Q3+Q4+Q5 = 3678.4+13320+16744+90400+1447.2 = 125589.6 J