Question

Q.17 A Carnot engine whose low temperature reservoir is at 7°C has an efficiency of 50%. It is desired to increase the efficiency to 70%. By how many degrees should the temperature of the high temperature reservoir be increased
(1) 840 K
(2) 280 K
(3) 560 K
(4) 380 K​

Answer 1

Explanation:

Hia buddy

here's your answer

we know that the formula,

$n = \frac{t1 - t2}{t1}$

n is given as 50% = 50/100 = 0.5

t2 is given 7°C = 280°K

$0.5 = \frac{t1 - 280}{t1}$

So from here t1 = 560°K

now,

n = 70% = 70/100 = 0.7

t2 = 280°K

So again using formula

$0.7 = \frac{t1 - 280}{t1}$

t1' = 933°K

Now we have to find ∆T = 933 - 560 = 380°K

So your answer is 380°K

Hope it helps uh!

Answer 2

Explanation:

☣️PLEASE MARK AS BRAINLIEST☣️