A carnot engine takes 2000J of heat from a reservoir at 500K does some work and discards some heat to a reservoir at 350K.How much heat is discarded,how much work does the engine do and what is the efficiency ? Physics class 12 numericals

Physics with Akhtiar Ali  Bullo

A carnot engine takes 2000J of heat from a reservoir at 500K does some work and discards some heat to a reservoir at 350K.How much heat is discarded,how much work does the engine do and what is the efficiency ? Physics class 12 numericals  Inter class physics numericals New sindh text physics numericals  12 class physics numerical solution  Second law of thermodynamics numericals solution

Transcript

00:00Numerical number 1 of unit 17 second law of thermodynamics is a Carnot engine takes 2000

00:16joules of heat from a reservoir 8500 Kelvin does some work and discords some heat to a

00:26reservoir 8 350 Kelvin how much heat is discarded how much work does the engine do and what is the efficiency

00:41so this problem may have a car a heat engine a Joe 2000 joules heat

00:48there a year from a reservoir or get a temperature pay 500 Kelvin pay so who conquer a force or heat

00:59co release career to a low reservoir at which temperature 350 Kelvin so we have to find q2

01:09means heat released or discarded and also work done and also efficiency so sab se peli hum data banayenge

01:27we have obtained q1

01:29which is 2000 joules and also we have obtained t1 means temperature of hot reservoir which is 500 Kelvin

01:50and we have to find heat lost or discarded means q2

01:54and also we have to calculate efficiency

02:04and we have obtained t2 means temperature of low reservoir 350 Kelvin and also we have to calculate

02:17work done so for solution or calculation

02:24first of all we will find efficiency as efficiency is calculated by using formula eta is equal to

02:361 minus t2 upon t1 into 100 percent

02:47so now by putting the values

02:49the values efficiency will be equal to 1 minus t2 is obtained as 350 Kelvin divided by

03:03t1 which is 500 Kelvin into 100 percent

03:09so eta will be equal to 0.7 into 100 percent so eta will be equal to 1 minus 350 divided by 500 it will be 0.7 into 100 percent

03:26so efficiency will be obtained as jab 1 mein se 0.7 subtract karvayenge so we will get

03:350.3 into 100 percent so efficiency will be obtained as 0.3 into 100 it will be 30 percent

03:52so this is our first answer

03:59now we will calculate q2

04:02as we know that q2 upon q1 is equal to t2 upon t1

04:19&

04:36normal

04:37so here we have to come to

04:39q1 divide

04:41side multiply so q2 will be equal to t2 upon t1 enter q1 so q2 will be equal to as t2 is obtained

05:00as 350 kelvin divided by t1 which is 500 kelvin into q1 q1 is also obtained which is 2000 joules

05:16so q2 will be equal to 350 divided by 500 it will be 0.7 into 2000 so q2 will be obtained as 0.7 into

05:382000 it will be 1400 joules so this is our second answer now finally we will calculate work done

05:56as work done is calculated by using formula w is equal to q1 minus q2 in terms of heat engine or

06:07carno engine so work done will be equal to as q1 is given which is 2000 joules minus q2 is also obtained

06:22which is 1400 joules so we will work done obtained hooga jab 2000 mein se 1400 subtract karvayenge

06:34so we will get 600 joules so this is our third answer

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