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Heat Transfer Question

  • #1
103
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Homework Statement



A piece of metal with a mass of 1.50 kg, specific heat of 200 J/kg*oC, and initial temperature of 100oC is dropped into an insulated jar that contains liquid with a mass of 3.00 kg, specific heat of 1,000 J/kg*oC, and initial temperature of 0oC. The piece of metal is removed after 5 seconds, at which time its temperature is 20oC. Neglect any effects of heat transfer to the air or insulated jar.

The temperature of the liquid after the metal is removed is ... ?


Homework Equations


[delta]U = Q + W


The Attempt at a Solution



I'm not sure how to apply an equation to solve this. Any guidance?
 
Last edited:

Answers and Replies

  • #2
103
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The options for the temperature of the liquid after the metal is removed is:
0oC
4oC
8oC
10oC
20oC
 
  • #3
Redbelly98
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Homework Equations


[delta]U = Q + W

The Attempt at a Solution



I'm not sure how to apply an equation to solve this. Any guidance?
Hello science.girl,

Hmmm, that equation you wrote doesn't seem to apply here.

Do you have your physics textbook or class notes handy? You're looking for an equation that contains the specific heat, mass, and temperature change in it -- often denoted by c, m, and ΔT, respectively.
 
  • #4
103
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Hello science.girl,

Hmmm, that equation you wrote doesn't seem to apply here.

Do you have your physics textbook or class notes handy? You're looking for an equation that contains the specific heat, mass, and temperature change in it -- often denoted by c, m, and ΔT, respectively.
The only equation I can think of is

c = Q/(m[Tf - Ti])

But, I'm not given a value Q. I would be looking for Tf, right?
 
  • #5
103
0
Any help? Is there an equation that I'm missing?
 
  • #6
142
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Any help? Is there an equation that I'm missing?

Q=mc*(change in temperature)

Where: Q absorbed = Q released

So you can set a side equal to each part of the equation then plug in...

m*c*(Tf-Ti) for the liquid = m*c(Tf-Ti) for the metal
 
  • #7
103
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Q=mc*(change in temperature)

Where: Q absorbed = Q released

So you can set a side equal to each part of the equation then plug in...

m*c*(Tf-Ti) for the liquid = m*c(Tf-Ti) for the metal
Now I see how the equation is applied. Thank you so much for your help. :smile:
 
  • #8
142
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Now I see how the equation is applied. Thank you so much for your help. :smile:
no prob :biggrin:
 
  • #9
2,017
85
Also before you do any problem in thermo like this use the 1st law and reduce it down, its good practise to start doing this as it helps to visualise the problem. These are relatively simple but if you ever do anything with the steady flow energy equation you'll see why doing this is useful.

so in this case you take

dU = dQ + qW

we know there is no work, but there is heat transfer.

dU = dQ

This means that the heat flow from the bar into the liquid is the only thing to affect the internal energy. so this means we can apply the Q=mcdt as above.
 
  • #10
103
0
Also before you do any problem in thermo like this use the 1st law and reduce it down, its good practise to start doing this as it helps to visualise the problem. These are relatively simple but if you ever do anything with the steady flow energy equation you'll see why doing this is useful.

so in this case you take

dU = dQ + qW

we know there is no work, but there is heat transfer.

dU = dQ

This means that the heat flow from the bar into the liquid is the only thing to affect the internal energy. so this means we can apply the Q=mcdt as above.
Interesting. This is very useful, Chris. Thank you for this explanation... I will try to apply it to my further studies of physics! :smile:
 

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