Finding specific heat of an object graphically

  • Thread starter armolinasf
  • Start date
  • #1
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Homework Statement



Samples A and B are at different initial temperatures when they are placed in a thermally isolated container and allowed to come to thermal equilibrium. Figure a gives their temperatures T versus time t. Sample A has a mass of 5.2 kg; sample B has a mass of 1.6 kg. Figure b is a general plot about the material of sample B. It shows the temperature change T that the material undergoes when energy is transferred to it as heat Q. The change T is plotted versus the energy Q per unit mass of the material.




The Attempt at a Solution



The heat gained by object b must equal the total heat of object a. So Qb=Qa=Q/mb*mb, mb=mass of object b.

but Q=mcΔt ==> c=Q/mΔt would Δt then just be the total change from 100 to 60 degrees?

this would give me 25600/(5.2*60)=82.05 which is incorrect. Where Am I going wrong?

Thanks for the help
 

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Answers and Replies

  • #2
993
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The heat gained by object b must equal the total heat of object a.
I think it is better to replace the underlined by 'heat lost by'.
 
  • #3
993
13
but Q=mcΔt ==> c=Q/mΔt would Δt then just be the total change from 100 to 60 degrees?
I think that you mean that [itex]\Delta[/itex]T is the change from 100 to 40, i.e. 60deg because we are considering A.
 
  • #4
993
13
So Qb=Qa=Q/mb*mb, mb=mass of object b.
Can the poster explain what the above mean?
 

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