A Carnot engine uses a steam boiler at 100°C as the high-temperature reservoir. The low-temperature reservoir is the outside environment at 35.0°C. Energy is exhausted to the low-temperature reservoir at a rate of 150 W. Hint (a) Determine the efficiency of the engine. (Give your answer as a percentage.) eC = % (b) Determine the useful power output of the heat engine (in W). output = W (c) How much steam (in kg) will it cause to condense in the high-temperature reservoir in 2.00 h? condensate = kg

(a) η = 0.1742 = 17.42 %

(b) W = 31.64 W

(c) m = 0.58 kg

Explanation:

(a)

The efficiency of a Carnot's engine is given as follows:

where,

η = efficiency = ?

T₁ = source temperature = 100°C + 273 = 373 K

T₂ = sink temperature = 35°C + 273 = 308 K

Therefore,

η = 0.1742 = 17.42 %

(b)

Another formula for the efficiency of Carnot's Engine is:

where,

Q₁ = Input heat rate

Q₂ = Heat rejected = 150 W

Therefore,

Q₁ = 181.64 W

Now, for the useful power output:

where,

W = Useful Output

W = 31.64 W

(c)

The heat required to condense steam i given as:

where,

m = mass of steam condensed = ?

t = time = 2 h = 7200 s

H = Latent heat of condensation of steam = 2260 KJ/kg = 2260000 J/kg

Therefore,

m = 0.58 kg

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete