The heat capacity of a system at constant volume is given by $C_v=Z T^2$ where Z = 0.045 J/K³

A system is originally at 250 K, and a thermal reservoir at 125 K is available. Determine the maximum amount of work that can be recovered as the system is cooled down to the temperature of the reservoir.

Question

The heat capacity of a system at constant volume is given by [latex]C_v=Z T^2[/latex] where Z = 0.045 J/K³

A system is originally at 250 K, and a thermal reservoir at 125 K is available. Determine the maximum amount of work that can be recovered as the system is cooled down to the temperature of the reservoir.

Accepted Answer

Refer Fig. 5.48.

Initial temperature of the system, [latex]T_1=250[/latex] K

Temperature of the reservoir, [latex]T_2=125[/latex] K

Heat capacity of the system at constant volume is

[latex]C_v=Z T^2=0.045[/latex] T²

[latex]\left[\begin{array}{l} ext { The product of mass and specific heat }(m c) ext { is called the } \ ext { heat capacity of the substance. The capital letter } C, C_p \ ext { or } C_v ext { is used for heat capacity }\end{array} ight][/latex]

Heat removed from the system

[latex]Q_1=\int_{T_1}^{T_2} C_v d T=\int_{250}^{125} 0.045 T^2 d T[/latex]

= [latex]0.045\left[\frac{T^3}{3} ight]_{250}^{125}=\frac{0.045}{3}\left(125^3-250^3 ight)[/latex]

= [latex]-205.08 imes 10^3[/latex] J

[latex](\Delta S)_{ ext {system }}=\int_{250}^{125} C_v \frac{d T}{T}=\int_{250}^{125} 0.045 T^2 \frac{d T}{T}[/latex]

= [latex]0.045 \int_{250}^{125} T d T=0.045\left[\frac{T^2}{2} ight]_{250}^{125}[/latex]

= [latex]\frac{0.045}{2}\left(125^2-250^2 ight)=-1054.7[/latex] J/K

[latex](\Delta S)_{ ext {reservoir }}=\frac{Q_1-W}{T_{ ext {reservoir }}}=\frac{205.08 imes 10^3-W}{125}[/latex] J/K

[latex](\Delta S)_{ ext {working fluid in } H E}=0[/latex]

∴ [latex](\Delta S)_{ ext {universe }}=(\Delta S)_{ ext {system }}+(\Delta S)_{ ext {reservoir }}=-1054.7+\frac{205.08 imes 10^3-W}{125}[/latex]

Since [latex](\Delta S)_{ ext {universe }} \geq 0[/latex]

∴ [latex]-1054.7+\frac{205.08 imes 10^3-W}{125} \geq 0[/latex]

or [latex]-1054.7+1640.64-\frac{W}{125} \geq 0 \quad ext { or } \quad 585.94-\frac{W}{125} \geq 0[/latex]

or [latex]585.94 \geq \frac{W}{125} \quad ext { or } \quad \frac{W}{125} \leq 585.94[/latex]

i.e., [latex]W_{(\max )}=585.94 imes 125= 7 8 . 2 4[/latex] kJ.

Question 4.45: The heat capacity of a system at constant volume is given by...

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