Rohith Gandhi
Last Activity: 15 Years ago
Dear Rohan,
case 1
4kg of P at 60 degree and 1 kg of R at 50 degree, resultant temp of mix = 55 degree
so, energy lost by P = energy gained by R and from calorimetry,
4XSpX(60-55)=1XSrX(55-50) =>Sr = 4Sp
case 2
1kg of P at 60 degree and 1 kg of Q at 50 degree, resultant temp of mix = 55 degree
so, energy lost by P = energy gained by Q and from calorimetry,
1XSpX(60-55)=1XSqX(55-50) =>Sq = Sp
from both above cases, we get, Sr = 4Sq
case 3
1kg of Q at 60 degree and 1 kg of R at 50 degree, resultant temp of mix = ?, let it be T
so, energy lost by Q = energy gained by R and from calorimetry,
1XSqX(60-T)=1XSrX(T-50) => SqX(60-T)=4SqX(T-50) =>60-T = 4T - 200 =>5T = 260 => T = 260/5 = 52 degree
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Regards,
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Rohith Gandhi