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
00:24เคฏเคนเคพเค เคชเคฐ เคนเคฎเฅเค Q2 means heat loss calculate เคเคฐเคจเฅ เคนเฅ เคคเฅ เคธเคฌเคธเฅ เคชเคนเคฒเฅ เคนเคฎ data เคฌเคจเคพเคเคเคเฅ
00:35is power is given which is 500 kilowatt
00:50as we know that 1 kilowatt is equal to 1000 watt so เคเคธเคเฅ เคนเคฎ watt เคฎเฅเค convert เคเคฐเฅเคเคเฅ
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 เคญเฅ เคนเคฎเฅเค obtained เคนเฅ 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 เคญเฅ เคนเคฎเฅเค obtained เคนเฅ 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
02:36เคฏเคนเคพเค เคชเคฐ Q1 เคญเฅ เคนเคฎเฅเค obtained เคจเคนเฅเค เคนเฅ เคเคฐ work เคญเฅ obtained เคจเคนเฅเค เคนเฅ เคคเฅ เคธเคฌเคธเฅ เคชเคนเคฒเฅ เคนเคฎ
02:46เคฏเคน find เคเคฐเฅเคเคเฅ so first we will find work done เคฏเคนเคพเค เคชเคฐ power or time given เคนเฅ so in terms of power เคนเคฎ work เคเฅ find เคเคฐเฅเคเคเฅ
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 เคเคฐเฅเคเคเฅ 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
04:16เคเคฌ เคนเคฎ Q1 เคเฅ เคฏเคนเคพเค เคฒเฅเคเฅ เคเคพเคเคเคเฅ เคคเฅ multiply เคเคฐเฅเคเคพ เคเคฐ eta means efficiency เคฏเคนเคพเค multiply เคเคฐ เคฐเคนเฅ เคนเฅ เคคเฅ เคฏเคนเคพเค เคชเคฐ เคเคเคฐ divide เคเคฐเฅเคเฅ
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 เคเคฐเฅเคเคเฅ
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
06:49you
06:50you
06:50you
06:52you
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