What is the final temperature of the mixture?

  • Thread starter Thread starter Byeongok
  • Start date Start date
  • Tags Tags
    Thermodynamics
AI Thread Summary
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.
Byeongok
Messages
13
Reaction score
0

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
 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top