Question

How much heat is required to convert 10.0 kg of ice from a temperature of - 20 0C to water vapor at a temperature of 213 F?

Answer 1

Useful information:
heat of fusion of water = 334 J/g
heat of vaporization of water = 2257 J/g
specific heat of ice = 2.09 J/g·°C
specific heat of water = 4.18 J/g·°C
specific heat of steam = 2.09 J/g·°C

m = 10 kg = 10000 g

Step 1: Heat required to raise the temperature of ice from -20 °C to 0 °C Use the formula

q1 = mcΔT = 10000*2.09* ( 0--10) = 209000 J

Step 2: Heat required to convert 0 °C ice to 0 °C water

q2 = m·ΔH_f =   10000*334 =3340000 J

Step 3: Heat required to raise the temperature of 0 °C water to 100 °C water

q3 = mcΔT = 10000*4.18* ( 100-0) = 4180000 J

Step 4: Heat required to convert 100 °C water to 100 °C steam

q4 = m·ΔH_v =  10000*2257 = 22570000 J

Step 5: Heat required to convert 100 °C steam to 100.555556 °C steam

213 F =100.555556°C

q5 = mcΔT = 10000*2.09* ( 100.555556- 100) =11611.1204​ J

Heat_Total = Heat_{Step 1} + Heat_{Step 2} + Heat_{Step 3} + Heat_{Step 4} + Heat_{Step 5}

= 209000+ 3340000+4180000+22570000+11611.1204

= 30310611.1 J =3.03106111e7 J