Water Temperature mixing together

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SUMMARY

The equilibrium temperature when mixing 1 kg of water at 100°C with 10 kg of water at 0°C is slightly above 0°C. The calculation involves using the formula Q = mcΔT, where Q represents heat, m is mass, c is specific heat capacity (4186 J/kg°C for water), and ΔT is the change in temperature. The heat lost by the hot water equals the heat gained by the cold water, leading to a final temperature of approximately 9.09°C. This scenario exemplifies a weighted average problem in thermodynamics.

PREREQUISITES
  • Understanding of thermodynamics principles
  • Familiarity with the specific heat capacity of water (4186 J/kg°C)
  • Knowledge of the Q = mcΔT formula
  • Basic algebra for solving equations
NEXT STEPS
  • Study the concept of heat transfer in thermodynamics
  • Learn about latent heat and its effects on phase changes
  • Explore weighted averages in mathematical problems
  • Investigate the specific heat capacities of different substances
USEFUL FOR

Students studying physics or chemistry, educators teaching thermodynamics, and anyone interested in understanding heat transfer and equilibrium temperature calculations.

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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