A heat engine takes in 6.45 × 103 J of thermal energy from a reservoir at 600 K

Q: A heat engine takes in 6.45 × 103 J of thermal energy from a reservoir at 600 K

Q1 = 6.45*10^3 , n =0.45 , T1 =600 K n = 1- (Q2/Q1) 0.45 = 1- (Q2/(6.45*10^3)) Q2 = 3.547*10^3 J Q1/Q2 = T1/T2 6.45/3.547 = 600/T2 T2 =330 K T1 = 315 K , T2 = 280 K, m =1 kg , Cp = 4186 J/kg.K , DT = 1K Q1 = mCP*DT = 1*4186*! =4186 J Q2/Q1 = T2/T1 Q2/4186 = 280/315 Q2 =3720.9 j W =Q1 -Q2 = 4186 -3720.9 W = 465 J

ID: 1444929 • Letter: A

Question

A heat engine takes in 6.45 × 103 J of thermal energy from a reservoir at 600 K and returns some of this energy to a reservoir at TL < 600 K .If this engine operates at an efficiency of 0.450, what is the maximum value possible for TL?

A heat engine operates on a Carnot cycle that runs clockwise between a reservoir at 315 K and a reservoir at 280 K. One cycle moves enough energy from the high-temperature reservoir to raise the temperature of 1.0 kg of water by 1.0 K. How much work is done by the engine in one cycle?

Explanation / Answer

Q1 = 6.45*10^3 , n =0.45 , T1 =600 K

n = 1- (Q2/Q1)

0.45 = 1- (Q2/(6.45*10^3))

Q2 = 3.547*10^3 J

Q1/Q2 = T1/T2

6.45/3.547 = 600/T2

T2 =330 K

T1 = 315 K , T2 = 280 K, m =1 kg , Cp = 4186 J/kg.K , DT = 1K

Q1 = mCP*DT = 1*4186*! =4186 J

Q2/Q1 = T2/T1

Q2/4186 = 280/315

Q2 =3720.9 j

W =Q1 -Q2 = 4186 -3720.9

W = 465 J

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A heat engine takes in 6.45 × 103 J of thermal energy from a reservoir at 600 K

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