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10 litres of hot water at 70^@C is mixed...

`10` litres of hot water at `70^@C` is mixed with an equal volume of cold water at `20^@ C`. Find the resultant temperature of the water. (Specific heat of water `= 4200 J//kg -K`).

Text Solution

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Resultant temperature, `theta = (m_(1)S_(1)theta_(1)+m_(2)S_(2)theta_(2))/(m_(1)S_(1)+m_(2)S_(2))`
Here, `m_(1) = m_(2) = 10 kg`
(since mass = `1` litre of water is `1 kg`)
`theta_(1) = 70^@ C, theta_(2) = 20^@ C`
and `S_(1) = S_(2) = 4200 J//kg -K`
`theta = (10xx4200 xx 70 +10 xx 4200 xx 20)/(10 xx4200 +10xx 4200) = 45^@ C`.
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