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:[tex]= \dfrac{5 \ kg }{1590 \ kg/m^3}[/tex]
= 0.0031 m³
Suppose [tex]\beta[/tex] is independent of temperature while pressure is constant;
Then:
The change in volume can be expressed as:
[tex]\int ^{V_2}_{V_1} \dfrac{dV}{V} =\int ^{T_2}_{T_1} \beta dT[/tex]
[tex]In ( \dfrac{V_2}{V_1}) = \beta (T_2-T_1)[/tex]
[tex]V_2 = V_1 \times exp (\beta (T_2-T_1))[/tex]
[tex]V_2 = 0.0031 \ m^3 \times exp (1.2 \times 10^{-3} \times 20)[/tex]
[tex]V_2 = 0.003175 \ m^3[/tex]
However; the workdone = -PdV
[tex]W = -1.01 \times 10^5 \ Pa \times ( 0.003175 m^3 - 0.0031 \ m^3)[/tex]
W = - 7.6 J
The heat energy Q = Δ h
[tex]Q = mC_p(T_2-T_1)[/tex]
[tex]Q = 5 kg \times 0.84 \ kJ/kg^0 C \times 20[/tex]
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
