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A metal piece of 50 g specific heat 0.6 ...

A metal piece of 50 g specific heat 0.6 cal/`g^(@)C` initially at `120^(@)`C is dropped in 1.6 kg of water at `25^(@)`C. Find the final temperature or mixture.

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`m_(1)c_(1)(theta_(1)-theta) " " = m_(2)c_(2)(theta-theta_(2))`
`=m_(2)c_(2)(theta-theta_(2))`
`:. " " c_(2)=1 cal//gm^(@)C`
`:. " " 50xx0.6xx(120-theta)=1.6xx10^(3)xx1xx(theta-25)`
`theta = 26.8^(@)C`
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