Question

(a) How much energy is needed to raise 1.00 kg of liquid water from 19.9 ?C to 99.9 ?C? (b) How much energy is needed to evaporate 1.00 kg of liquid water (at 100?C) into steam (at 100?C)? Be careful not to confuse J with kJ. (c) Is the relative size of these two numbers consistent with everyday experience that it takes much longer (e.g. maybe an hour or so) to boil away an entire pot of water than it does (e.g. maybe 10 minutes) to heat the water up from room temperature to boiling temperature?

Answer 1

Part A

Energy required to raise the temperature of water is given by

Q1 = m*C*dT

m = 1 kg

C = specific heat of water = 4186 J/kg-C

dT = 99.9 - 19.9 = 80

Q1 = 1*4186*80 = 334880 J

Q1 = 0.335*10^6 J

Part B

Energy required for evaporation is given by:

Q2 = m*Lv

Lv = Latent heat of vaporization = 2.26*10^6 J/kg

m = 1 kg

Q2 = 1 kg*(2.26*10^6 J/kg)

Q2 = 2.26*10^6 J

Part C

from part A and B we can see that Q2 > Q1

Now Suppose if we want to heat 1 kg water on a 500 W heater, then time taken will be

for Q1

P = Q1/t

t = Q1/P = 0.335*10^6/500 = 670 sec = 11.17 min

for Q2

t = Q2/P = 2.26*10^6/500 = 4520 sec = 75.33 min

So we can see that these numbers are consistent with everyday experience.