A reversible engine take in heat from a reservoir of heat at 285°C, 35minute and give out heat to sink at 105°C, 18 minute.How many calories per second must it take from the reservoir in order to produce useful work at the rate of 308j/s
How?
To find the rate at which the reversible engine must take heat from the reservoir, we can use the formula for the efficiency of a heat engine:
Efficiency = Useful work output / Heat input
The efficiency (η) of a Carnot engine, which is a reversible engine, can be expressed as:
η = 1 - (T_cold / T_hot)
Where T_hot is the temperature of the heat source (in Kelvin) and T_cold is the temperature of the heat sink (in Kelvin).
Given that the heat source has a temperature of 285°C, we can convert this to Kelvin by adding 273:
T_hot = 285 + 273 = 558 K
Similarly, the heat sink has a temperature of 105°C:
T_cold = 105 + 273 = 378 K
Substituting these temperatures into the efficiency equation:
η = 1 - (378 / 558) = 1 - 0.678 = 0.322
We know that the useful work output is 308 J/s. The heat input can be calculated using the efficiency:
Efficiency = Useful work output / Heat input
0.322 = 308 / Heat input
Heat input = 308 / 0.322 = 956.52 J/s
To convert this to calories per second, we use the conversion factor:
1 calorie = 4.184 J
Therefore, the rate at which the reversible engine must take heat from the reservoir is:
956.52 J/s * (1 calorie / 4.184 J) = 228.6 calories/s
So, the reversible engine must take 228.6 calories per second from the reservoir in order to produce useful work at the rate of 308 J/s.