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3. (20 points) Consider a heat engine that produces power at a rate of 50 Watts with an efficiency of 20%. This engine takes heat from a flame at 1400 K and rejects heat to flowing water in a heat exchanger at 300 K a. What heat input rate is required by this engine? b. What water flow rate is required such that the temperature in the flowing liquid water doesn't increase by more than 10 °C? c. If a Carnot heat engine were operated between the same temperature reservoirs and provided the same power output of 50 W, what heat input rate would be required?

          3. (20 points) Consider a heat engine that produces power at a rate of 50 Watts with an efficiency of 20%.
This engine takes heat from a flame at 1400 K and rejects heat to flowing water in a heat exchanger at
300 K
a. What heat input rate is required by this engine?
b. What water flow rate is required such that the temperature in the flowing liquid water doesn't
increase by more than 10 °C?
c. If a Carnot heat engine were operated between the same temperature reservoirs and provided
the same power output of 50 W, what heat input rate would be required?
        
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3. (20 points) Consider a heat engine that produces power at a rate of 50 Watts with an efficiency of 20%.
This engine takes heat from a flame at 1400 K and rejects heat to flowing water in a heat exchanger at
300 K
a. What heat input rate is required by this engine?
b. What water flow rate is required such that the temperature in the flowing liquid water doesn't
increase by more than 10 °C?
c. If a Carnot heat engine were operated between the same temperature reservoirs and provided
the same power output of 50 W, what heat input rate would be required?

Added by Josefina C.

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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300K. What heat input rate is required by this engine? increase by more than 10°C? For the same power output of 50 W, what heat input rate would be required?
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Transcript

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00:01 So we can say for part a, we know that the efficiency epsilon is going to be defined by the work done, divided by the heat, the total heat intake.
00:10 We can say that here we know the temperature of the hot reservoir would essentially be equal to 100 degrees celsius or we can say 373 kelvin.
00:23 The temperature of the cold reservoir would be equal to 60 degrees celsius and this would be equal to 3 .3 .5.
00:31 33 kelvin.
00:33 At this point we can find the maximum efficiency of the engine.
00:37 The efficiency would then be equal to t sub h minus t sub l divided by t sub h.
00:44 So this would be 373 minus 333 over 373.
00:56 This is going to equal 0 .107 or we can say 107...
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