Question

A heat engine aborbs 2KJ of heat from a reservoir at 150°C and rejects 0.5 Kj of heat to a sink in one complete cycle. Calculate :
i) Efficiency of the Engine.
ii)Temperature of the sink.
iii) Work done by the heat engine during each cycle.
iv) Efficiency of the engine if the temperature of the sink is further decreased by 50K.​

Answer 1

Answer:

Learn from the attachment

Answer 2

Answer: $T_{2} = -167^{o}\ C, \eta = 75\%, W = 1500J and \eta' = 88\%$

Explanation:

Given that,

$Q_{1} = 2000J \\</p><p></p><p>Q_{2} = 500J \\</p><p></p><p>T_{1} = 150^{o}\ C+273 = 423K$

We know that,

(I). The temperature of the sink is

$\dfrac{Q_{2}}{Q_{1}} = \dfrac{T_{2}}{T_{1}} \\</p><p></p><p>T_{2} = \dfrac{Q_{2}}{Q_{1}}\times T_{1} \\</p><p></p><p>T_{2} = \dfrac{500\ J}{2000\ J}\times 423\ K \\</p><p></p><p>T_{2} = 106K -273 = -167^{o} C$

(II). The efficiency of the engine is

$\eta = 1 - \dfrac{T_{2}}{T_{1}} \\</p><p></p><p>\eta =1 - \dfrac{106\ K}{423\ K} \\</p><p></p><p>\eta = 0.75 = 75\%$

(III). Work done by the heat engine during each cycle

$W = Q_{1} - Q_{2} \\</p><p></p><p>W = 2000J-500J \\</p><p></p><p>W = 1500J$

(IV). Efficiency of the engine if the temperature of sink is further decreased by 50 K

$\eta'= 1-\dfrac{T_{2}}{T_{1}} \\</p><p></p><p>\eta' = 1-\dfrac{50K}{423K} \\</p><p></p><p>\eta' = 0.88 = 88\%$

Hence, this is the required solution