Answer:
(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:
[tex]\eta = 1-\frac{T_2}{T_1}[/tex]
where,
η = efficiency = ?
T₁ = source temperature = 100°C + 273 = 373 K
T₂ = sink temperature = 35°C + 273 = 308 K
Therefore,
[tex]\eta = 1 - \frac{308\ k}{373\ k} \\[/tex]
η = 0.1742 = 17.42 %
(b)
Another formula for the efficiency of Carnot's Engine is:
[tex]\eta = 1 - \frac{Q_2}{Q_1}[/tex]
where,
Q₁ = Input heat rate
Q₂ = Heat rejected = 150 W
Therefore,
[tex]0.1742 = 1 - \frac{150\ W}{Q_1}\\\\ \frac{150\ W}{Q_1} = 1-0.1742\\\\Q_1 = \frac{150\ W}{0.8258}[/tex]
Q₁ = 181.64 W
Now, for the useful power output:
[tex]W = Q_1-Q_2\\W = 181.64\ W - 150\ W\\W = 31.64\ W[/tex]
where,
W = Useful Output
W = 31.64 W
(c)
The heat required to condense steam i given as:
[tex]Q = mH\\Q_1t=mH\\m = \frac{Q_1t}{H}[/tex]
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,
[tex]m = \frac{(181.64\ W)(7200\ s)}{(2260000\ J/kg)}[/tex]
m = 0.58 kg
