How much heat is required to raise the temperature of 12.0 g of water from − 35 °C to 175 °C? heat: kcal

Answer 1

The process you describe has 5 separate pieces as shown in the graph below:

Section A is the heating of the solid ice from -35 to zero.

Section B is the phase change of solid ice to liquid water.

Section C is the heating of liquid water from zero to 100.

Section D is the phase change from liquid water to gaseous steam

Section E is the heating of steam from 100 to 185.

Section A: We use the equation Q = mC_iΔT where Ci is the specific heat of ice (2.06 j/(g°C) and ΔT = 35 °C

So Q = (12)(2.06)(35) = 865.2 joules

Section B: We use Q = mΔH_f where ΔH_f is the heat of fusion or the heat required to accomplish the phase change of ice to water (334.16 J/g). So Q = (12)(334.16) = 4009.92 joules.

Section C: Again we use the equation Q = mC_wΔT but this time Cw is the specific heat of liquid water (4.184 j/(g°C) and ΔT = 100 °C

So Q = (12)(4.184)(100) = 5020.8 joules

Section D: Again we use Q = mΔH_v where ΔH_v is the heat of vaporization or the heat required to accomplish the phase change of water to steam (2259 J/g). So Q = (12)(2259) = 27108 joules.

Section E: Again we use the equation Q = mC_SΔT but this time CS is the specific heat of steam (2.02 j/(g°C) and ΔT = 85 °C

So Q = (12)(2.02)(85) = 2060.4 joules

So the total heat required is 865.2 + 4009.92 + 5020.8 + 27108 + 2060.4 = 39064.32 joules. Converting to kCal we get (39064.32 joules)(0.000239006 kcal/1 joule) = 9.34 kcal