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Question
If water of mass 60 g and temperature 60 ˚C is mixed with water of mass 60 g and temperature 40 ˚C, what will be the maximum temperature of the mixture?
Solution
Let maximum common temperature of mixture be T ˚C
According principle of heat exchange,
heat lost by hot water = heat gained by cold water
mHot c (THot - T) = mCold c (T - TCold)
Given: mHot = mCold = 60 g
∴ T = `("T"_"Hot" + "T"_"Cold")/2 = (60 + 40)/2` = 50 ˚C
Thus, maximum temperature of mixture will be 50 ˚C.
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