Problem 2.
A nonconducting container filled with 25 kg of water at 20°C is fitted with a stirrer, which is made to
turn by gravity acting on a weight of mass 35 kg. The weight falls slowly through a distance of 5 m in
driving the stirrer. Assuming that all work done on the weight is transferred to the water and that the
local acceleration of gravity is 9.8 m∙s−2, determine:
(a) The amount of work done on the water.
(b) The internal energy change of the water.
(c) The final temperature of the water, for which CP = 4.18 kJ∙kg−1∙°C−1.
(d) The amount of heat that must be removed from the water to return it to its initial temperature.
(e) The total energy change of the universe because of (1) the process of lowering the weight, (2) the
process of cooling the water back to its initial temperature, and (3) both processes together.
Solution
Part a: The work done is equal to change in potential energy of the 35 kg mass.
𝑊 = ∆(𝑚𝑔𝑧) = 35 kg x 9.8 m/s2 x 5 m = J (not in J/kg)
Part b: ∆𝑈 𝑡 = 𝑄 + 𝑊
𝑄 = 0, so ∆𝑈 𝑡 is calculated as equal to 𝑊.
∆𝑈 𝑡 = 𝑊 = ∆𝑚𝑈 = 𝑚∆𝑈
𝑚 is here the mass of water, i.e., 25 kg.
Part c: ∆𝐻 = ∆𝑈 + ∆(𝑃𝑉), H=U+PV
In case of water (system), there is no change in pressure and volume. Therefore ∆(𝑃𝑉) = 0.
𝑇2
∆𝑈 = ∆𝐻 = ∫ 𝐶𝑃 𝑑𝑇
𝑇1
It is given that 𝐶𝑃 is a constant value 4.18 kJ∙kg−1∙°C−1. Therefore:
𝑚∆𝑈 = 𝑚∆𝐻 = 𝑚𝐶𝑃 (𝑇2 − 𝑇1 )
So only unknown is 𝑇2 .
Part d: Now we will apply the first law from state 1 the higher temperature state, to the final state 2, the
lower temperature state. There is no work done in this case.
So: ∆𝑈 𝑡 = 𝑄 + 𝑊; but 𝑊 = 0, therefore ∆𝑈 𝑡 = 𝑄
𝑄 = 𝑚∆𝑈 = 𝑚∆𝐻 = 𝑚𝐶𝑃 (𝑇2 − 𝑇1 ). But here 𝑇2 is 20oC (293.15 K), and 𝑇1 is the higher temperature
(i.e. the 𝑇2 of Part c). So numerically, the value will be same as in Parts a and b, but sign will be negative.
Part e: (∆Energy of system + ∆Energy of surroundings) = 0 (A)
Universe = System + Surroundings
So: (∆Energy of system + ∆Energy of surroundings) = ∆Energy of universe (B)
Comparing Eqs (A) and (B): ∆Energy of universe = 0
Therefore in all the three cases of Part e, answer is 0.
Problem 2.8
A closed, nonreactive system contains species 1 and 2 in vapor/liquid equilibrium. Species 2 is a very
light gas, essentially insoluble in the liquid phase. The vapor phase contains both species 1 and 2. Some
additional moles of species 2 are added to the system, which is then restored to its initial T and P. As a
result of the process, does the total number of moles of liquid increase, decrease, or remain
unchanged?
Solution y1=0.3, y1=0.5,
y1=0.5, y2=0.7 y2=0.5
y2=0.5
x1≈1, x2≈0 x1≈1
x1≈1
N = 2, 𝜋 = 2
F = 2-𝜋+N = 2 – 2 + 2 = 2.
As the two variables are chosen here as T and P, which are to be kept same. Therefore, x1, x2, y1, and y2
are set by the system itself and cannot be chosen at our will.
Because the addition of species 2 is only going to the vapor phase, so it will change the composition of
the vapors.
In order to regain the initial values of y1 and y2, some of the liquid must evaporate. Thus the moles of
the liquid decrease.
Problem 2.18
Liquid water at 180°C and 1002.7 kPa has an internal energy (on an arbitrary scale) of 762.0 kJ⋅kg−1 and
a specific volume of 1.128 cm3⋅g−1.
(a) What is its enthalpy?
(b) The water is brought to the vapor state at 300°C and 1500 kPa, where its internal energy is 2784.4
kJ⋅kg−1 and its specific volume is 169.7 cm3⋅g−1. Calculate ΔU and ΔH for the process.
Solution
Part a: H=U+PV
H= 762.0 kJ/kg + 1002.7 kPa x 1.128 cm3/g x 1000 g/kg x 1 bar/100 kPa x 1 J/ 10 cm3-bar x 1 kJ/1000 J
H=
Part b: The value of H in Part a is now H1.
∆𝑈 = 𝑈2 − 𝑈1 = 2784.4 – 762.0
𝐻2 = 𝑈2 + 𝑃2 𝑉2
= 2784.4 + 1500 kPa x 169.7 cm3/g x 1000 g/kg x 1 bar/100 kPa x 1 J/ 10 cm3-bar x 1 kJ/1000 J
Then calculate, ∆𝐻 = 𝐻2 − 𝐻1 .
See; there is no need of temperature in the calculations.
Mass Balance for Open Systems
∆ Mass of System + ∆ Mass of Surroundings = 0.
Time rate of change of mass of control volume + net time rate of mass of flowing streams through the
control surface = 0
𝑑𝑚cv /𝑑𝑡 + ∆(𝑚̇)fs = 0
∆(𝑚̇)fs = 𝑚̇3 − 𝑚̇1 − 𝑚̇2
We know that: 𝑚̇ = 𝜌𝑢𝐴
𝑑𝑚cv /𝑑𝑡 + ∆(𝜌𝑢𝐴)fs = 0 (Continuity Equation)
𝑑𝑚cv
Under steady-state conditions (SSC): The accumulation (or depletion) term, = 0.
𝑑𝑡
Therefore: ∆(𝜌𝑢𝐴)fs = 0
SSC does not mean steady-flow (SF), i.e. it does not mean that the flow-rates are constant. It only means
that the sum of all input flow-rates is equal to the sum of all output flow-rates.
For single input and output: 𝜌2 𝑢2 𝐴2 − 𝜌1 𝑢1 𝐴1 = 0
𝜌2 𝑢2 𝐴2 = 𝜌1 𝑢1 𝐴1
If flow-rate is constant, then the process is also called steady-flow (SF). Then 𝑚̇ = constant.
Then:
𝜌2 𝑢2 𝐴2 = 𝜌1 𝑢1 𝐴1 = 𝑚̇ = const.
Another form:
𝑢2 𝐴2 𝑢1 𝐴1 𝑢𝐴
= = = 𝑚̇
𝑉2 𝑉1 𝑉
All the examples
Problems 22, 23, 25-37, 40, 41.
