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