Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.750 kg of water decreased from 125 °C to 63.5 °C.

Answer 1

These are the thermodynamic values for water that I have obtained. Your values may differ slightly.

∆Hvap = 2260 J/g

C(steam) = 2.09 J/gº

C(liquid) = 4.184 J/gº

step1: bring water/steam from 125º to 100º

q = mC_steam∆T = (750 g)(2.09 J/gº)(25º) = 39,188 J = 39.2 kJ

step 2: phase change from steam @100º to liquid @100º

q = m∆Hvap = (750 g)(2260 J/g) = 1,695,000 J = 1695 kJ

step 3: reduce temperature of liquid from 100º to 63.5º

q = mC_liquid∆T = (750 g)(4.184 J/gº)(63.5º) = 119,558 J = 119.6 kJ

Summing all the heat values we have...

39.2 kJ + 1695 kJ + 119.6 kJ = 1854 kJ = 1850 kJ (3 sig. figs.) = energy change (note: all heat values would be negative since heat is being lost upon cooling from 125º to 63.5º)