A heat engine performs work of 0.4166 watts in one hour and rejects 4500J of heat to the sink.What is the efficiency of engine ?

Physics with Akhtiar Ali  Bullo

A heat engine performs work of 0.4166 watts in one hour and rejects 4500J of heat to the sink.What is the efficiency of engine ? Physics class 12 numericals  inter class physics numerical solution  12 class physics numericals  second law of thermodynamics numericals  New sindh text physics numericals  #physics

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00:00numerical number six of unit 17 second law of thermodynamics is a heat engine performs

00:12work of 0.4166 watts in one hour and rejects 4500 joules of heat to the sink what is the

00:25efficiency of engine here we call a heat engine work of 0.4116 watts in one hour and how much

00:41heat rejects 4500 heat rejects to the sink so we find efficiency in engine

00:49so sb se phele hem data banayenge is is here per joh zero point four one six six

01:05say ye power ka value hai so power is obtained is zero point four one six six watts or time is also given

01:21which is one hour is one hour is equal to thirty six hundred seconds

01:31or q2 is also given which is forty five hundred joules and we have to find

01:43efficiency

01:50yahan par hum efficiency ko calculate karenge in terms of heat energy

01:55so for solution

01:59or calculation we will use formula of efficiency as eta is equal to one minus q2 upon q1

02:11multiplied with 100 percent

02:18yahan par q1 ka jo value hai woh hume obtained nahi hai so first of all we will find q1

02:26is work done is equal to q1 minus q2

02:33so q1 will be equal to work done plus q2

02:40so q1 will be equal to work done plus q2

02:42so q1 will be equal to work done plus q2

02:44and q1 minus po2 minus woh hume Build desd nous

02:52as we know that

02:55is power is equal to work done plus power unit time

02:59time so work done will be equal to power enter time so work done will be equal to

03:09is power is given which is 0.4166 watts multiplied with time which is also given

03:193600 seconds so work done will be obtained as when we will multiply these

03:30values then we will get final answer is 1499.76 joules so we have obtained the

03:43value of work done now we will calculate q1 so q1 will be equal to work done plus q2 so q1 will be

03:56equal to is work done is also obtained which is 1499.76 plus q2 q2 is given which is 4500 joules

04:11so when we will simplify these values means add these values then we will get final answer of q1

04:22is 5999.76 joules so this is the answer for q1 now finally

04:39finally we will calculate efficiency so efficiency is equal to 1 minus q2 upon q1 multiplied with a

04:59hundred percent so efficiency will be equal to 1 minus is the value of q2 is given which is 4500 joules

05:13divided by q1 the value of q1 is 5999.76 multiplied with a hundred percent so efficiency will be obtained as

05:29when we will be told then we will tribate the value of c1 by uh

05:400.006 about q1 of q1 from 1 among q2 multiplied is 1 minus c1 1 of q1 by 7

05:46down the value of q1 from californiaались 0.75 from r501 we will get 0.75 multiplied by

05:52percent so efficiency will be obtained is 0.25 into 100 it will be 25 percent so

06:04this is our final answer for efficiency

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