Question

The following information is given for water at 1 abm boiling point 100.00 C AHap100.00 C)-2.259x10 Jig AH(0.000 °C)-333.5 Jg melting point 0.000 C specific heat gas 2.010 J/g specific heat liquid-4.184 g C C A 34.70 g sample of liquid water is instially at 69.10 C. If the sample is heated at constant pressure (P-1 atm), kJ of enerpy are needed to raise the temperature of the sample to 119.20 C

Answer 1

Given weight of the liquid water = 34.7 grams

Initial temperature is 69.10^oC

The sample has to be heated to 119.20^oC

The specific heat of water is different than that of specific heat of water vapor. So we need to make two different calculations those are upto 100^oC and beyond 100^oC

1. Specific heat of liquid water is 4.184 J.g^-1.^oC^-1

The amount of energy required to raise the temperature can be calculated with the formula

Q = mC$\Delta$T

m is the mass of liquid water = 34.7 grams

C is the specific heat of water = 4.184 J.g^-1.^oC^-1

$\Delta$T is change in temperature = 100 - 69.1^oC = 30.9^oC

Therefore,

Q = 34.7 grams x 4.184 J.g^-1.^oC^-1 x 30.9^oC = 4447.6 Joules.

Thus the energy needed to take 34.7 grams of water from 69.1 to 100^oC is 4447.6 Joules

2. At 100^oC, the liquid water converts into water vapor. For this phase transition from one state to another state, some amount of heat is required which is given by heat of vaporization $\Delta$ H_vap

Given $\Delta$ H_vap for water is 2.259 x 10³ J/g

This energy is to covert 1 gram of water into 1 gram of water vapor at 100^oC

Therefore $\Delta$ H_vapfor 34.7 grams of water is

34.7 x 2.259 x 10³ Joules = 78387.3 Joules

Thus the energy needed to convert 34.7 grams of liquid water to vapor at 100^oC is 78387.3 Joules

3. Now, water is in the form of vapor. Specific heat of water vapor given is 2.010 J.g^-1.^oC^-1

Thus the amount of heat required to raise the temperature of 34.7 grams of water vapor from 100^oC to 119.20^oC is

Q = mC$\Delta$T

Q = 34.7 grams x 2.010 J.g^-1.^oC^-1 x (119.20 - 100^oC) = 1339.1 Joules.

Thus the energy needed to raise the temperature of 34.7 grams of water from 100 to 119.2^oC is 1339.1 Joules

Sum of the three calculated heats give the amount of heat required to raise the temperature of 34.7 grams of water from 69.1 to 119.2^oC i.e.,

4447.6 Joules + 78387.3 Joules + 1339.1 Joules = 84,174 J = 84.174 kJ (1 kJ = 1000 Joules)

Thus the required amount of energy in kJ is 84.174 kJ