Please Help Me - Clausius-Clapeyron Equations?

Question

How much heat (in kJ) is required to convert 423 g of liquid H2O at 24.0°C into steam at 152°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C . The heat of vaporization (ΔHvap) is 40.65 kJ/mol.)

Jeremiah Adams · Accepted Answer

Heat Required# = 1135.5kJ#

Calm down.. take these problems ONE step at a time, and they become simple arithmetic.

FIRST - warm the water. We have liquid water, so the first heat required is to increase it from #24.0^oC# to boiling at #100.0^oC#. For this we will use the heat capacity of liquid water, #4.184 J/(g·°C)#. Make sure the units match up, and calculate.

#423g xx (100 - 24) xx 4.184 J/(g·°C) = 134507J#

Or, #134.5kJ#

SECOND - vaporize the water. For this we use the heat of vaporization: #40.65 "kJ"/"mol"#.

#423g xx "1 mol"/("18 g H"_2"O") xx 40.65 = 955.3kJ#

THIRD - heat the steam using the heat capacity of steam, 2.078 J/g·°C, #423g xx (152 - 100) xx 2.078 J/(g·°C) = 45708J#

Or, #45.7kJ#

FINALLY, add them all up!

Heat Required = #134.5 + 955.3 + 45.7 = 1135.5kJ#

Peyton Adams · Answer

The total energy required is 1135.5 kJ

There are three separate events occurring here, and each must have its energy calculated separately: Water is warmed from 24.0 °C to 100 °C: This calculation is: #q=mC_sDeltaT = 423g xx4.184J/(g °C) xx 76 °C= 134507 J=134.5 kJ# Next, the water must be converted to steam at 100 °C. Since we have the molar heat of vapourization, we need to note that 423 g is 23.5 moles (#423-:18#) #q=23.5 "mol" xx 40.65 (kJ)/"mol" = 955.3 kJ# Finally, the steam is heated from 100 °C to 152 °C #q= 423 xx 2.078 J/(g °C) xx 52 °C = 45708 J = 45.7 kJ# So, all together, the energy required is #134.5+955.3+45.7 = 1135.5kJ#