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bagirrra123 [75]
3 years ago
15

A real heat engine operates between temperatures Tc and Th. During a certain time, an amount Qc of heat is released to the cold

reservoir. During that time, what is the maximum amount of work Wmax that the engine might have performed? Express your answer in terms of Qc, Th, and Tc.
Physics
1 answer:
Vlad [161]3 years ago
5 0

Answer:

W_{max} =Q_C(\dfrac{T_H}{T_C} - 1)

Explanation:

given,

Temperature of heat engine operate between

Th (temperature in hot reservoir) and Tc(temperature in cold reservoir)

amount of heat released to = Qc

to find maximum amount of work = ?

now,

efficiency of heat engine

\eta = 1 - \dfrac{T_C}{T_H} = 1 -\dfrac{Q_C}{Q_H}

now,

\dfrac{T_C}{T_H} = \dfrac{Q_C}{Q_H}

Q_H= \dfrac{T_H}{T_C} Q_C

 maximum work =

W_{max} = Q_H - Q_C

W_{max} =\dfrac{T_H}{T_C} Q_C - Q_C

W_{max} =Q_C(\dfrac{T_H}{T_C} - 1)

above expression gives the expression of maximum amount of work.

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Read 2 more answers
The three stages of a train route took 1 hour ,2 hours ,and 4 hours . The first two stages were 80km and 200km of the train aver
luda_lava [24]

Answer:

the third stage was 480 km long

Explanation:

Stage 1:

Time = 1 hours

Speed = 80km

Stage 2:

Time =  2 hours

Speed = 200km

Stage 3:

Time =  4 hours

Let the Distance at the stage 3 be x

Average speed of the train route = 100 km/h

So

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 0

\frac{ \text{speed at stage 1} + \text{speed at stage 2} + \text{speed at stage 3}}{3} = 100

Lets find the speed at stage 1

Speed =  \frac{Distance }{Time}

Speed =  \frac{80}{1}

Speed 1= 80 km/hr

The speed at stage 2

Speed =  \frac{Distance }{Time}

Speed =  \frac{200}{2}

Speed 2  = 100 km/hr

The speed at stage 3

Speed =  \frac{Distance }{Time}

Speed =  \frac{x}{4}

Speed 3  = \frac{x}{4}

we kow that average is ,

\frac{ \text{speed 1} + \text{speed 2} + \text{speed 3}}{3} = 100

\frac{ 80 + 100+ \frac{x}{4} }{3} = 100

\frac{ 180 + \frac{x}{4} }{3} = 100

\frac{ \frac{720 +x}{4} }{3} = 100

\frac{720 +x}{4} \times \frac{1}{3} = 100

\frac{720 +x}{12} = 100

720 +x = 100 \times 12

720 +x = 1200

x = 1200- 720

x = 480

6 0
3 years ago
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