How much heat is required to raise the temperature of 1kg of lead from 25⁰C  to 100⁰C.Specific heat capacity of lead=0.128 J/g⁰C Physics class 12 numericals solution  Inter class physics numerical solution  First law of thermodynamics numericals - video Dailymotion

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How much heat is required to raise the temperature of 1kg of lead from 25⁰C  to 100⁰C.Specific heat capacity of lead=0.128 J/g⁰C Physics class 12 numericals solution  Inter class physics numerical solution  First law of thermodynamics numericals  12 class physics numericals

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00:00Numerical number 8 of unit 16 first law of thermodynamics is how much heat is required

00:13to raise the temperature of 1 kilogram of lead from 25 degree centigrade to 100 degree

00:21centigrade. Specific heat capacity of lead is given which is 0.128 Jol upon gram into centigrade.

00:32So we have to find heat required means delta Q is equal to what mass of lead which is

00:51given as 1 kilogram which is also equal to 1000 gram and initial temperature T1 is given

01:03as 25 degree centigrade. Final temperature T2 is given as 100 degree centigrade and specific

01:17heat capacity of lead is obtained as 0.128 Jol upon gram into degree centigrade. For calculation

01:36or solution we will use formula is specific heat is equal to delta Q upon m delta T. So

01:48delta Q will be equal to mc delta T. So delta Q will be equal to mc delta T. So delta Q will

02:04be equal to is mass is given as 1000 into specific heat which is also obtained as 0.128 Jol upon gram

02:16into degree centigrade into degree centigrade into degree centigrade into delta T. So delta T is equal

02:22to T2 minus T1. So delta Q will be equal to 1000 into 0.128 Jol upon t2 which is 100 degree centigrade

02:41minus T1 minus T1 which is 25 degree centigrade. So delta Q will be equal to is 1000 into 0.128. So it will be 128 into

02:48128 into 100 minus 25 so it will be 75. So delta Q will be obtained as 128 into 75. So it will be 90,

03:02to 100 minus 25, so it will be 75. So, delta Q will be obtained as 128 into 75, so it will

03:15be 9600 Joule. So, this is our final answer. The answer in book is given as 0, 2, 4, 0,

03:32Joule. So, this is incorrect.

How much heat is required to raise the temperature of 1kg of lead from 25⁰C to 100⁰C.Specific heat capacity of lead=0.128 J/g⁰C Physics class 12 numericals solution Inter class physics numerical solution First law of thermodynamics numericals

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