To develop the problem we will start by finding the energy taken by each cycle through the efficiency of the motor and the exhausted energy. Later the work will be found for the conservation of energy in which this is equivalent to the difference between the two calculated energy values. Finally the estimated time will be calculated with the work and the power given,
[tex]\text{Efficiency of the heat engine} = \eta = 26\% = 0.26[/tex]
[tex]\text{Energy taken in by the heat engine during each cycle} = Q_h[/tex]
[tex]\text{Energy exhausted by the heat engine in each cycle} = Q_c = 8.55*10^3 J[/tex]
[tex]\eta = 1 - \frac{Q_{c}}{Q_{h}}[/tex]
[tex]0.26 = 1 - \frac{8.55\ast 10^{3}}{Q_{h}}[/tex]
[tex]\frac{8.55* 10^{3}}{Q_{h}} = 0.74[/tex]
[tex]Q_h = \frac{8.55*10^3}{0.74}[/tex]
[tex]Q_h = 11.554*10^3J[/tex]
PART A)
Work done by the heat engine in each cycle = W
[tex]W = Q_h-Q_c[/tex]
[tex]W = 11.554*10^3J-8.55*10^3J[/tex]
[tex]W = 3004J[/tex]
According to the value given we have that,
[tex]P = 4.0kW[/tex]
[tex]P = 4000W[/tex]
Power is defined as the variation of energy as a function of time therefore,
[tex]P = \frac{W}{t}[/tex]
[tex]4000W = \frac{3004J}{t}[/tex]
[tex]t = \frac{3004}{4000}[/tex]
[tex]t = 0.75s[/tex]
Therefore the interval for each cycle is 0.75s
