Find the temperature of Liquid A after 12 minutes

  • #1
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Homework Statement


A liquid A initially at temperature T1= 28.4°C, is heated by an immersion heater in container. The resulting temperature T2 is recorded at 1 minute intervals and the following results obtained.
(I will post a picture)
(a) Complete the table by computing temperature changes ΔT=T2-T1
I've done this and the missing values are: 7.6, 14.7, 24.1, 32.1,40.6,46.7,55.9 degrees Celsius.
(b) Plot a graph of ΔT against time (t). I've also done this.
(c) Find the slope S of the graph, I've also done this and I got 8.33.
(d) Cp the Specific Heat capacity of the liquid is related to the slope S by : Cp=2000/S J Kg-1K-1.
Find the Specific Heat Capacity of the Liquid.
I've also done this and got 240.1J Kg-1K-1.
(e) What would be the temperature of the liquid after 12 minutes?
This is where I'm stuck.

Homework Equations


Eh=m*C*ΔT

The Attempt at a Solution


I'm really stuck, I was thinking now I have the specific heat capacity i could find the energy supplied, then relate that to the temperature change but I don't have the mass. They basically want X°C-28.4°C and that would give you the temperature change after 12 minutes then simply add to the initial temperature to get the final.

But I'm pretty unsure I've been racking my brain looking for formulas linking temperature and time.
 

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Answers and Replies

  • #2
TSny
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(e) What would be the temperature of the liquid after 12 minutes?
This is where I'm stuck.
In part (c) you didn't include the units for the slope. This should help with answering part (e). Interpret what the slope is telling you.
 
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  • #3
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The unit would then be degrees Celsius per minute?
 
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TSny
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  • #5
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Wait, so it's as simple as doing 8.33 degrees celsius per min *12 +28.4 to give me 128.36°C
 
  • #6
TSny
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Wait, so it's as simple as doing 8.33 degrees celsius per min *12 +28.4 to give me 128.36°C
Yes, that looks right. However, I don't know how you got your value of 8.33 Co/min for the slope. It appears to be a little too large.
 
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On My graph I drew a line of best fit, so I suppose there was human error involved in finding the gradient.
 
  • #8
TSny
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On My graph I drew a line of best fit, so I suppose there was human error involved in finding the gradient.
Ok. I think your work looks good.
 

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