#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# A reversible engines operates with an efficiency of 50% if during each cycle it rejects 150 cals to a reservoir of heat at 30C  then,1)what is the temperature of the other reservoir?2)how much work does it carry out per cycle?

arun
123 Points
5 years ago
efficiency$(\eta )$ is 50% = 0.5
$\eta = 1-\frac{T(sink)}{T(source)}$
$0.5 = 1-\frac{30}{T(source)}$
$0.5 = \frac{30}{T(source)}$
$T(source)=60$ degree celcius
$\eta = 1-\frac{Q(sink)}{Q(source)}$
$0.5 = 1-\frac{150}{Q(source)}$
$Q(source)=300$ calories
$work= Q(source) - Q(sink)$
$work= 150$ calories
so temprature of other reservoir is 60 degree celcius
and work done by engine in one cycle is 150 calories