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a 50.0-g sample of liquid water at 25.0 degrees C is mixed with 29.0 g of water at 45.0 degrees C.

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What is the final Temperature of the water mixture

Apr 12th, 2015

Heat balance is:  heat gained by 50g sample = heat lost by 29 g water

So m1 x cp x temp. Diff. Of 1 = m2 x cp x temp diff of 2

m1 x temp. Diff of 1 = m2 x temp diff of 2

Let t be the final temp of water mixture.

50 g ( t-25 ) C = 29 g ( 45 -t) C

50 t -1250 = 1305 -29 t

Therefore t = 2555/79  = 32.34 C

Answer :  final temp. Is 32.34 C

Apr 12th, 2015

I have answered your question.

Niru

Apr 12th, 2015

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