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Three liquids of the same amount are mix...

Three liquids of the same amount are mixed. The specific heats of these liquids are `s_(1),s_(2) and s_(3)` and their initial temperature are `theta_(1),theta_(2) and theta_(3)`, respectively. Find out the final temperature of the mixture.

Text Solution

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Let the final temperature of the mixture be `theta` and the mass of each liquid be m.
`therefore` Heat absorbed by the 1st liquid, `H_(1)=ms_(1)(theta-theta_(1))`
Heat absorbed by the 2nd liquid, `H_(2)=ms_(2)(theta-theta_(2))`
Heat absorbed by the 3rd liquid, `H_(3)=ms_(3)(theta-theta_(3))`
Since no heat is supplied from outside,
`therefore" "H_(1)+H_(2)+H_(3)=0" ...(1)"`
[any one or two of `H_(1),H_(2),H_(3)` must be negative, implying heat lost, then equation (1) means, heat gained = heat lost]
`therefore" "ms_(1)(theta-theta_(1))+ms_(2)(theta-theta_(2))+ms_(3)(theta-theta_(3))=0`
`"or, "theta(s_(1)+s_(2)+s_(3))=s_(1)theta_(1)+s_(2)theta_(2)+s_(3)theta_(3)`
`"or, "theta(s_(1)+s_(2)+s_(3))=s_(1)theta_(1)+s_(2)theta_(2)+s_(3)theta_(3)`
`"or, "theta=(s_(1)theta_(1)+s_(2)theta_(2)+s_(3)theta_(3))/(s_(1)+s_(2)+s_(3))`
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