Question

For each engine calculate ΔE = QH−Wout−QC, where QH is the amount of heat transferred from the hot reservoir, QC is the amount of heat transferred to the cold reservoir, Wout is the energy output of the heat engine. QH, Wout, QC are positive quantities.

Answer 1

Answer:

The energy output of the of the heat engine at various location is  0J, -1J and 0J

Explanation:

The change in internal energy of the process is according to the diagram

    $\delta E =dQ-dW$

    $\delta E_{a} = 50-20-30$

          = 0 J

     $\delta E_{b} = 10-7-4$

            = -1 J

    $\delta E_{c} = 30-20-10$

          = 0 J

Answer 2

The given equation is an expression of the first law of thermodynamics,

that states that the energy in a closed system is conserved.

The values for each engine are;

(a) ΔE = 0 J

(b) ΔE = -1 J

(c) ΔE = 0 J

Reasons:

Given equation; ΔE = $Q_H$ - $W_{out}$ - $Q_C$

Required:

To calculate ΔE = $Q_H$ - $W_{out}$ - $Q_C$ for the three heat engines;

Solution:

(a) $Q_H$ = 50 J, $W_{out}$ = 30 J, $Q_C$ = 20 J

Therefore;

ΔE = 50 J - 30 J - 20 J = 0 J

(b) $Q_H$ = 10 J, $W_{out}$ = 4 J, $Q_C$ = 7 J

Therefore;

ΔE = 10 J - 4 J - 7 J = -1 J

(b) $Q_H$ = 30 J, $W_{out}$ = 10 J, $Q_C$ = 20 J

Therefore;

ΔE = 30 J - 10 J - 20 J = 0 J

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