A carnot engine absorbs 52kJ as heat and exhaust 36kJ as heat in each cycle. Calculate: (a) The engine efficiency (b) The work done per cycle in kilojoules Physics class 12 numericals Inter class physics numericals 12 class physics numerical solution New Sindh Text Physics numericals New Sindh board physics numericals Second law of thermodynamics numericals
00:00Numerical number 8 of unit 17 second law of thermodynamics is a Carnot engine absorbs
00:1552 kilo joules is heat and exhaust 36 kilo joules is heat in each cycle calculate a the engine efficiency
00:32b the work done per cycle in kilo joules
00:38Numerical number is a Carnot engine which is 52 kilo joules is heat absorb and 36 kilo joules is release in each cycle
00:52so we will need efficiency to find a and b work done to find each cycle
01:04so we will make data first
01:12is heat absorbed means q1 is given which is 52 kilo joules
01:23and heat released means q2 is also given which is 36 kilo joules
01:33A we will find efficiency in engine and b we will find work done
01:42find find find like in kilo joules
01:44for solution
01:49first we find efficiency mean solution for a
01:59here we find efficiency
02:02we find the formula we use
02:06so efficiency will be equal to
02:091 minus q2 upon
02:12q1
02:14multiplied with 100 percent
02:17so efficiency will be equal to
02:221 minus
02:24as the value of q2 is
02:2636 kilojoules
02:28divided by
02:30q1
02:33the value of q1 is also given
02:35which is 52 kilojoules
02:37into 100 percent
02:39here we will not convert
02:43the value of q2 is equal to 1 minus
02:51when we will divide 52 from 36 then we will get 0.692 multiplied with 100 percent
02:57so efficiency will be obtained is when we will subtract 0.692 from 1 then we will get 0.308 and 100 percent will remain same
03:12so efficiency will be obtained is when we will multiply 100 with 0.308 then we will get 30.8 percent
03:26so this is our answer for efficiency so this is our answer for efficiency now solution for b means work done so work done w will be calculated by using formula q1 minus q2
03:43q1 minus q2 so work done will be obtained is is the value of q1 is given is 52 kilojoules minus the value of q2 is given is 36 kilojoules
04:03so when we will subtract 36 from 52 then we will get 16 kilojoules
04:18so when we will subtract 36 from 52 then we will get 16 kilojoules so this is our answer for work done
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