Question

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature changes from 0∘C to 20∘C0 ∘ C to 20 ∘ C Determine ΔVt,W,Q,ΔHt, and ΔUt.ΔV t ,W,Q,ΔH t , and ΔU t . The properties for liquid carbon tetrachloride at 1 bar and 0∘C0 ∘ C may be assumed independent of temperature: β=1.2×10−3K−1,CP=0.84kJ⋅kg−1⋅K−1, and rho=1590kg⋅m−3β=1.2×10 −3 K −1 ,C P =0.84kJ⋅kg −1 ⋅K −1 , and rho=1590kg⋅m −3

Answer 1

Answer:

Explanation:

From the information given:

Mass of carbon tetrachloride = 5 kg

Pressure = 1 bar

The given density for carbon tetrachloride = 1590 kg/m³

The specific heat of carbon tetrachloride =  0.84 kJ/kg K

From the composition, the initial volume of carbon tetrachloride will be:$= \dfrac{5 \ kg }{1590 \ kg/m^3}$

= 0.0031 m³

Suppose $\beta$ is independent of temperature while pressure is constant;

Then:

The change in volume can be expressed as:

$\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT$

$In ( \dfrac{V_2}{V_1}) = \beta (T_2-T_1)$

$V_2 = V_1 \times exp (\beta (T_2-T_1))$

$V_2 = 0.0031 \ m^3 \times exp (1.2 \times 10^{-3} \times 20)$

$V_2 = 0.003175 \ m^3$

However; the workdone = -PdV

$W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)$

W = - 7.6 J

The heat energy Q = Δ h

$Q = mC_p(T_2-T_1)$

$Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20$

Q = 84 kJ

The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;

ΔU = ΔQ + W

ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)

ΔU = 83.992 kJ