Asked by tobby
A 500 g piece of silver at 150 degree Celsius is submerged in 1000 g of water at 5 degree Celsius to be cooled. determine the final temperature of silver and water given
Cwater = 4.18 x 10 ^3 J/kg degree Celsius
Csilver = 2.4 x 10^2 J/kg degree Celsius
Cwater = 4.18 x 10 ^3 J/kg degree Celsius
Csilver = 2.4 x 10^2 J/kg degree Celsius
Answers
Answered by
mahesh
Heat released by 0.5 kg silver=heat gained by 1 kg water upto final temp T.
Or 0.5(2.4x100)(150-T)=1(4.18x1000)(T-5)
Or 1.2(150-T)=41.8(T-5)
Or 180-1.2T=41.8T-209
Or 43T=389
Or T=9.05 deg Celsius
Or 0.5(2.4x100)(150-T)=1(4.18x1000)(T-5)
Or 1.2(150-T)=41.8(T-5)
Or 180-1.2T=41.8T-209
Or 43T=389
Or T=9.05 deg Celsius
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