|Nov17-04, 05:57 AM||#1|
Hi all, i often help out in the chemistry section, was wondering if you could help me with a question:
A vessel of thermal capacity 500 J/oC contains 0.5 kg of water at 50oC.
0.2 kg of water at 20oC is added and perfectly mixed with the hotter water.
In this process, heat is lost by the hotter water and vessel to the colder water, so the hotter water and vessel cool down and the colder water heats up.
We will call the final temperature of the mixture (and vessel) ‘T’.
(a) Write an expression for the heat lost by the hotter water and vessel in terms of T.
(b) Write an expression for the heat gained by the colder water in terms of T.
(c) If we assume that no heat is lost from the vessel to the surroundings, the heat lost by the hotter water and vessel must be the same as the heat gained by the colder water. So by putting the above two expressions equal to each other, find the value of T.
The answer= 42.86oC
I know the answer if there it is just how to find it out. Thank you
|Nov17-04, 09:06 AM||#2|
The thermal capacity of water is 4.1868 J/g/0C, so your 500 g of hot water as a thermal capacity of 500*4.1868 J/0C.
Loss of heat by container plus hot water: (500*4.1868 + 500)*(50-T) J
Gain of heat by cold water: 200*4.18*(T-20)
Since both heat quantities are equal you can find T.
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