a) For the thermal efficiency we have
[tex]\eta_{th} = \frac{Q_{out}}{Q_{in}} = \frac{|W|}{|Q_h|}\\\eta_{th} = \frac{|W|}{|W|+|Q_2|}[/tex]
With the previously values we know that
[tex]W=8kW[/tex] and [tex]Q_L = 1440/6kW[/tex] (convert the min to sec)
Replacing the values
[tex]\eta_{th}=\frac{8}{8+1440/6}=\frac{1}{31}\\\eta_{th}\% = 3.225\%[/tex]
b) We use the formula of carnot efficiency
[tex]\eta_{th}=1-\frac{T_l}{T_h}\\\eta_{th}\% =(1-\frac{280}{295})*100\\\eta_{th}\%=5.085\%[/tex]
**Note that apply the formula of carnot cycle we need to consider that there is no exchange of heat, there is no friction and the reservior are completely insulated
