Water Temperature mixing together

AI Thread Summary
When mixing 1 kg of water at 100°C with 10 kg of water at 0°C, the equilibrium temperature will be slightly above 0°C. The calculation involves using the formula Q = mcΔT to equate the heat lost by the hot water to the heat gained by the cold water. The heat lost by the 1 kg of water is 418,600 J, which equals the heat gained by the 10 kg of water, calculated as 41860 J multiplied by the final temperature. This results in a final temperature of approximately 9.09°C, confirming that it is indeed slightly above 0°C. The discussion highlights the importance of understanding heat transfer and weighted averages in thermal equilibrium problems.
strawberry7
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Homework Statement


When 1 kg of water at 100 C is mixed with 10 Kg of water at 0C, the equilibrium temperature will be:

a) exactly 0 C
b) Exactly 50 C
C) Exactly 100 C
d) slightly above 0 C

I think that the answer is D) sligtly above 0 C, but i don't know why. I'm not sure what formula to use, and I've been lookking all over the web and my textbook.
 
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When you add 1 kg of water in the liquid phase at 100 C, you can find the heat given out by it when its brought down to T c. That is equal to the heat gained by the 10 kg of water at 0c to come to Tc. Solve for T (final temperature).
 
so am i sopposed to use the Q= mc (Change in temperature) formula

so like 1 * 4186 * 100
Q = 418600And then 418600 = 10 *4186 *?
= 41860 * t
418600/41860 = t
t = 10

So it's slightly over 0 C ?
 
yea, intuitively you can see this is a weighted average kind of problem,
10 parts at 0, 1 part at 100, so ave for 11=9.09 degrees. Now if the water was ice, different animal as latent heats need to be included.
 
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