What is the final temperature of the mixture?

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In the experiment, 1.1 kg of aluminum at 92 degrees Celsius is mixed with 0.5 kg of water at 12 degrees Celsius to find the final temperature of the mixture. The heat exchange equation Qal = Qw is applied, leading to the equation 968(92 - t) = 2100(t - 12). The calculations reveal an error in arithmetic, resulting in an incorrect final temperature of 20.81 degrees Celsius. It is emphasized that checking calculations for mistakes is crucial in solving such problems. The discussion highlights the importance of accuracy in thermal equilibrium calculations.
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Homework Statement


In an experiment, 1.1kg of aluminium is heated to 92 degrees Celsius it is then dropped into 0.5kg of water at 12 degrees celsius.

Find the final temperature of the mixture. (Cw= 4200JKg-1Celsius-1) (Cal= 880JKg-1Celsius-1)

Homework Equations


Qal = Qw

The Attempt at a Solution


(1.1)(880)(92-t) = (0.5)(4200)(t-12)
968( 92 - t ) = 2100( t - 12 )
89056-968t = 2100t - 25200
63856 = 3068t
20.81 = t
 
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No.

As you suspected, I suppose.

You can check your answers yourself: if you put back "t=20.81" in the first line of your equation, you will see that you are way off.
From your second line, one can see that cooling and heating have a ratio of about 2:1 in temperature.
 
Last edited:
Byeongok said:

Homework Statement


In an experiment, 1.1kg of aluminium is heated to 92 degrees Celsius it is then dropped into 0.5kg of water at 12 degrees celsius.

Find the final temperature of the mixture. (Cw= 4200JKg-1Celsius-1) (Cal= 880JKg-1Celsius-1)

Homework Equations


Qal = Qw

The Attempt at a Solution


(1.1)(880)(92-t) = (0.5)(4200)(t-12)
968( 92 - t ) = 2100( t - 12 )
89056-968t = 2100t - 25200
63856 = 3068t
Your problem lies in the arithmetic used to get the line immediately above.

Always check your work for arithmetic mistakes.
 
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