A heat engine works at the rate of 500kW.The efficiency of the engine is 30%.Calculate the loss of heat per hour.

Physics with Akhtiar Ali  Bullo

A heat engine works at the rate of 500kW.The efficiency of the engine is 30%.Calculate the loss of heat per hour.  Physics class 12 numericals  Inter class physics numerical solution  12 class physics numericals  Second year physics numericals  New sindh text physics numericals  Second law of thermodynamics numericals

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00:00numerical number 5 of unit 17 second law of thermodynamics is a heat engine works at the

00:12rate of 500 kilowatt the efficiency of the engine is 30% calculate the loss of heat per hour

00:24यहां पर हमें Q2 means heat loss calculate करनी है तो सबसे पहले हम data बनाएंगे

00:35is power is given which is 500 kilowatt

00:50as we know that 1 kilowatt is equal to 1000 watt so इसको हम watt में convert करेंगे

00:59so 500 into 1000 watt so power will be equal to 5 lake watt or we can also write this is 5 into 10

01:24raised power 5 watt or efficiency भी हमें obtained है means eta is also given which is

01:3230% and we have to find loss of heat means Q2

01:44or time is also given time भी हमें obtained है that is 1 hour

01:55is 1 hour is equal to 3600 seconds for solution

02:09our calculation we will use formula to calculate Q2 is work is equal to Q1 minus Q2

02:26so Q2 will be equal to Q1 minus work done

02:36यहां पर Q1 भी हमें obtained नहीं है और work भी obtained नहीं है तो सबसे पहले हम

02:46यह find करेंगे so first we will find work done यहां पर power or time given है so in terms of power हम work को find करेंगे

03:02is power is equal to work done per unit time so work done will be equal to power into time

03:14so work done will be equal to is power is given which is 5 into 10 raised power 5 watt into time time is also given which is 30

03:32600 seconds so jaybham in ko simplify करेंगे then we will get the value of work done is

03:401.8 into 10 raised power 9 joules so we have get the value of work done

03:53now we will find Q1 is

04:02efficiency is also calculated by using formula work done upon heat supplied into 100%

04:14so we have to find Q1

04:16जब हम Q1 को यहां लेके जाएंगे तो multiply करेगा और eta means efficiency यहां multiply कर रही है तो यहां पर आकर divide करेगी

04:28so this equation will become is Q1 will be equal to work done upon efficiency into

04:35hundred percent

04:38so Q1 will be equal to is work done is obtained which is

04:471.8 into 10 is power 9 joules divided by efficiency which is also given

04:5530 percent multiplied with 100 percent

05:05percentage will be get cancelled with each other

05:09jab ham in values ko simplify करेंगे

05:12then we will get the final value of Q1 is

05:186 into 10 raised power 9 joules

05:256 into 10

05:32now finally we will find

05:37Q2

05:44is Q2 is equal to Q1 minus work done

05:51so Q2 will be equal to Q1 is 6 into 10 raised power 9 joules minus work done is given which is

06:041.8 into 10 raised power 9 joules

06:07so Q2 will be obtained as when we will subtract 1.8 from 6 then we will get 4.2 and 10 raised power 9 will remain same joules

06:26so the value of Q2 is obtained as 4.2 into 10 raised power 9 joules so this is our final answer

06:41so

06:43so

06:47you

06:49you

06:50you

06:52you

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