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
First, let's convert the temperature values from Celsius to Kelvin:
Reservoir temperature (T1) = 285°C + 273.15 = 558.15 K
Sink temperature (T2) = 105°C + 273.15 = 378.15 K
Next, we can calculate the heat transfer rates:
Rate of heat transfer from reservoir (Qin) = (285°C / 35 minutes) * (60 seconds/minute) = 164.57°C/minute
Rate of heat transfer to sink (Qout) = (105°C / 18 minutes) * (60 seconds/minute) = 350°C/minute
To find the heat transfer rate required to produce useful work, we can use the formula for efficiency:
Efficiency (η) = Useful work output / Heat input
308 J/s / Qin = η
Since it is a reversible engine, the efficiency can be expressed in terms of temperature:
η = 1 - (T2 / T1)
Substituting the efficiency equation, we can solve for Qin:
308 J/s = Qin * (1 - (T2 / T1))
308 J/s = Qin * (1 - (378.15 K / 558.15 K))
Qin = 308 J/s / (1 - (378.15 K / 558.15 K))
Calculating Qin:
Qin = 308 J/s / (1 - 0.6776)
Qin ≈ 941.58 J/s
To convert the heat transfer rate from joules per second to calories per second, we use the conversion factor:
1 calorie = 4.184 J
So, the heat transfer rate required in calories/second is:
Qin_calories = 941.58 J/s / 4.184 J/calorie
Qin_calories ≈ 225.05 calories/s
Therefore, the reversible engine must take approximately 225.05 calories per second from the reservoir to produce useful work at the rate of 308 J/s.