Final answer:
A reversible Carnot engine taking in heat at a rate needed to perform 750 W of work with a temperature difference between heat reservoirs at 527°C and 127°C would require approximately 358.2 cal/s. The correct option closest to this calculated value is option A: 357.1 cal/s.
Step-by-step explanation:
The student has asked how many calories per second must a reversible engine take from a reservoir at 527°C to perform work at the rate of 750W when it gives out heat to a sink at 127°C. To find the heat absorption rate from the heat reservoir by the Carnot engine, we'll need to use the efficiency formula of a Carnot engine, which is given by η = 1 - (Tc/Th), where Tc and Th are the absolute temperatures of the cold and hot reservoirs, respectively.
First, convert the Celsius temperatures to Kelvin (K = °C + 273.15):
Th = 527°C + 273.15 = 800.15 K
Tc = 127°C + 273.15 = 400.15 K
Now, calculate the efficiency of the engine:
η = 1 - (Tc/Th) = 1 - (400.15/800.15) = 1 - 0.5 = 0.5 or 50%
The power output of the engine is given in Watts, which equates to J/s, so 750 W is 750 J/s. Given the efficiency, the heat taken in from the reservoir in Joules per second is Qh = W/η = 750 J/s / 0.5 = 1500 J/s.
To convert this to calories per second (since 1 calorie = 4.184 J), divide by 4.184:
Qh(cal/s) = 1500 J/s / 4.184 J/cal ≈ 358.2 cal/s.
However, none of the options match this exactly; the closest to our calculated value is option A: 357.1 cal/s, which may be due to rounding differences or a typo in the question or the options provided.