A Carnot engine uses a steam boiler at 100°C as the high-temperature reservoir. The low-temperature reservoir is the outside environment at 16.5°C. Energy is exhausted to the low-temperature reservoir at a rate of 185 W.
(c) How much steam will it cause to condense in the high-temperature reservoir in 2.50 h?
condensate = ?kg
To determine the amount of steam that will condense in the high-temperature reservoir, we can use the formula for the efficiency of a Carnot engine:
efficiency = 1 - (T_low / T_high)
Where T_low is the temperature of the low-temperature reservoir in Kelvin and T_high is the temperature of the high-temperature reservoir in Kelvin.
In this case, the temperature of the low-temperature reservoir is 16.5°C, which is equivalent to 16.5 + 273.15 = 289.65 Kelvin. The temperature of the high-temperature reservoir is 100°C, which is equivalent to 100 + 273.15 = 373.15 Kelvin.
Using the efficiency formula:
efficiency = 1 - (289.65 / 373.15) = 0.2259
The efficiency of the Carnot engine is 0.2259, indicating that 22.59% of the energy is converted into useful work, while the remaining 77.41% is exhausted to the low-temperature reservoir.
Since the energy exhausted to the low-temperature reservoir is given as 185 W, we can calculate the total energy produced by the Carnot engine:
total energy = energy exhausted / efficiency = 185 / 0.2259 = 818.69 W
Now, we can calculate the amount of steam that will condense in the high-temperature reservoir using the heat energy gained as the basis. Assuming steam at 100°C is fully condensed to water:
latent heat of steam = 2260 kJ/kg
To convert the total energy from Watts to kilojoules:
total energy = 818.69 J/s * 3600 s = 2947.28 kJ
Now, we divide the total energy by the latent heat of steam to get the mass of steam:
mass of steam = total energy / latent heat of steam = 2947.28 / 2260 = 1.303 kg
Therefore, the amount of steam that will condense in the high-temperature reservoir in 2.50 hours is approximately 1.303 kg.