Answer:
the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg
Explanation:
Given that:
Initial Temperature [tex]T_1[/tex] = 15°C
Initial Pressure [tex]P_1[/tex] = 5 atm
Final Pressure [tex]P_2[/tex] = 1 atm
Data obtain from steam tables of saturated water at 15°C are as follows:
Specific volume v = 1.001 cm³/gm
The change in temperature = 2°C
Specific heat of water = 4.19 J/gm.K
volume expansivity β = 1.5 × 10⁻⁴ K⁻¹
The expression to determine the change in temperature can be given as :
[tex]\delta \ T = \frac{-V (1- \beta \ T}{C_p} * \delta \ P ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})[/tex][tex]\delta \ T = \frac{-1.001 \frac{cm^3}{gm} (1- 1.5*10^{-4} \ K^{-1} )*2}{4.19 \ \frac{J}{gm.K}} *(5-1)atm ( \frac{1}{9.87} \ \frac{J}{cm^3/atm})[/tex]
Δ T = 0.093 K
Now; we can calculate the lost work bt the formula:
[tex]W_{lost} = T_{surr} *S[/tex]
where ;
[tex]T_{surr}[/tex] is the temperature of the surrounding. = 20°C = (20+273.15)K = 293.15 K
From above the change in entropy is:
[tex]\delta \ S = C_p \ In (\frac{T+ \delta \ T }{T}) * \beta V \delta P[/tex]
[tex]\delta \ S = 4.19* \ In (\frac{288.15+0.093 }{288.15}) - 1.5*10^{-4} * 1.001 (5-1)* (\frac{1}{9.87})[/tex]
[tex]\delta \ S =1.408*10^{-3} \ J/gm.K[/tex]
[tex]W_{lost} = T_{surr} *S[/tex]
[tex]W_{lost} = 293.15* 1.408*10^{-3} \ J/gm.K[/tex]
[tex]W_{lost} = 0.413 \ kJ/kg[/tex]
Thus, the lost work per kilogram of water for this everyday household happening = 0.413 kJ/kg
