Answer:
[tex]W_{heater} = 849.54\,kJ[/tex]
Explanation:
The piston-cylinder system is modelled after the First Law of Thermodynamics:
[tex]W_{heater} - Q_{loss} + U_{1,sys} - U_{2,sys} + W_{1,b} - W_{2,b} = 0[/tex]
The electrical energy supplied by the resistance heater is:
[tex]W_{heater} = Q_{loss} + U_{2,sys} - U_{1,sys} + W_{2,b} - W_{1,b}[/tex]
Let suppose that air behaves ideally, so that:
[tex]W_{heater} = Q_{loss} + m\cdot c_{v} \cdot (T_{2}-T_{1}) + m\cdot P\cdot (\nu_{2}-\nu_{1})[/tex]
The initial specific volume is determined by the use of the equation of state for ideal gases:
[tex]P \cdot V = n \cdot R_{u}\cdot T[/tex]
[tex]P\cdot V = \frac{m}{M}\cdot R_{u}\cdot T[/tex]
[tex]P\cdot \nu = \frac{R_{u}\cdot T}{M}[/tex]
[tex]\nu = \frac{R_{u}\cdot T}{P\cdot M}[/tex]
[tex]\nu_{1} = \frac{\left(8.314\,\frac{kPa\cdot m^{2}}{kmol\cdot K}\right)\cdot (298.15\,K)}{(300\,kPa)\cdot (28\,\frac{kg}{kmol} )}[/tex]
[tex]\nu_{1} = 0.295\,\frac{m^{3}}{kg}[/tex]
The final specific volume can be derived from the following relationship:
[tex]\frac{\nu_{2}}{T_{2}} = \frac{\nu_{1}}{T_{1}}[/tex]
[tex]\nu_{2} = \frac{T_{2}}{T_{1}}\cdot \nu_{1}[/tex]
[tex]\nu_{2} = \frac{350.15\,K}{298.15\,K} \cdot (0.295\,\frac{m^{3}}{kg} )[/tex]
[tex]\nu_{2} = 0.346\,\frac{m^{3}}{kg}[/tex]
The energy supply is:
[tex]W_{heater} = 60\,kJ + (15\,kg)\cdot \left[\left(0.718\,\frac{kJ}{kg\cdot ^{\textdegree}C} \right)\cdot (77\,^{\textdegree}C- 25\,^{\textdegree}C) + (300\,kPa)\cdot \left(0.346\,\frac{m^{3}}{kg}-0.295\,\frac{m^{3}}{kg} \right)\right][/tex][tex]W_{heater} = 849.54\,kJ[/tex]
