How
much heat is required to raise the temperature of 7.0 g
of H2O from –20oC to 130oC?
There
are two phase changes, and three phases. Each phase must be
heated.
Step
1: The
ice must first be heated from –20oC to 0oC. [The
specific heats of ice and steam are both 0.5
cal/g/K.] Step 2: The temperature then remains constant
while the ice melts. The latent heat of fusion is 80 cal/g.
Step
3: The water is in the liquid phase as it is
heated from 0oC to 100oC. Step 4: Then 540 cal/g are required to vaporize
the water (but the temperature does not rise). Step 5: Finally, the steam must be heated from 100oC to 130oC.
The results are added, and the sum is multiplied by the quantity to get the total calories of heat required.
Step
1 warm
ice DQ/m
= cmDT = (0.5
cal/g/K)(20oC)
= 10 cal/g
Step 2 melt ice DQ/m = HLf = 80 cal/g
Step 3 warm water DQ/m = cmDT = (1.0 cal/g/K)(100oC) = 100 cal/g
Step 4 vaporize DQ/m = HLv = 540 cal/g
Step
5
warm
steam DQ/m
= cmDT = (0.5
cal/g/K)(30oC)
= 15 cal/g
_____________
Step 6 add 745 cal/g
Step
7
multiply
x 7.0 g
5215 cal
DQ is heat in
calories
m is mass in grams
cm is the specific heat in cal/g/K
DT is the
change in temperature within a specific phase
HLf is the latent heat of fusion in
cal/g
HLv is the latent heat of
vaporization in cal/g
