Specific heat capacity of a solid material

AI Thread Summary
To calculate the specific heat capacity of a solid material, a 4.80 kg sample is heated from 16.4°C to 219°C using 787 kJ of energy with an efficiency of 0.383. The formula used is c = Q / (m * ΔT), leading to an initial calculation of approximately 809.27 kJ/kg/K. However, due to the efficiency factor, the final specific heat capacity should be adjusted to reflect that only 38.3% of the energy effectively heats the material. Ultimately, the correct specific heat capacity is found to be 310 J/kg/K, confirming the calculations were accurate.
Aleisha
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Homework Statement


A 4.80 kg piece of solid material is heated from 16.4C to 219C (3 s.f.) using 787 kJ of energy (3 s.f.).

Assuming an efficiency of 0.383 for the heating process, and that the material does not melt, calculate the specific heat capacity of the material.

Homework Equations


Q=mc(Change in T)

The Attempt at a Solution


c=Q/ m (Delta T)
=(787e3)/ (4.80 x 202.6)
= 809.2711418 k/kg/K
However they have given me a efficiency of 0.383? Do i just time that by my answer? We haven't been taught about efficiency.

Thank you
 
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Aleisha said:
We haven't been taught about efficiency.
If you use 1kJ of energy in a heating process, but the process is only 30% efficient, how much heat ends up where it is wanted? (Not something that really requires teaching, I'd say.)
 
Your methodology is correct as is your answer if all of the heat energy goes into heating the body. If however, only 38.37% of the heat energy supplied goes into heating the body, you need to make an adjustment in your calculation.
 
haruspex said:
If you use 1kJ of energy in a heating process, but the process is only 30% efficient, how much heat ends up where it is wanted? (Not something that really requires teaching, I'd say.)
So my answer will be approximately 30% of 809.27k/kg/K?
 
Dr Dr news said:
Your methodology is correct as is your answer if all of the heat energy goes into heating the body. If however, only 38.37% of the heat energy supplied goes into heating the body, you need to make an adjustment in your calculation.
Thank you very much! So I should just find 38.3% of my answer (809.27)? Is that correct because it’s only being 30% efficient in its absorbance?
 
Aleisha said:
So my answer will be approximately 30% of 809.27k/kg/K?
That's the approach, but not 30%. That was a number I made up to illustrate. Use the given 0.383, i.e. 38.3%.
 
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haruspex said:
That's the approach, but not 30%. That was a number I made up to illustrate. Use the given 0.383, i.e. 38.3%.
Yes that’s understandable thank you :)
 
Aleisha said:
Yes that’s understandable thank you :)
There's another thread currently that looks the exact same problem, just slightly different numbers. I guess this is software generated. We can't find an error in the student's work (unless wrong numbers were posted in the thread), but the software is rejecting the answer.
I'd be interested to know whether you have better luck.
 
haruspex said:
There's another thread currently that looks the exact same problem, just slightly different numbers. I guess this is software generated. We can't find an error in the student's work (unless wrong numbers were posted in the thread), but the software is rejecting the answer.
I'd be interested to know whether you have better luck.
I got an overall answer of 310 J/kg/K and the answer was correct. All the data was copy and pasted from my question. Thank you for your help :)
 
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Aleisha said:
I got an overall answer of 310 J/kg/K and the answer was correct. All the data was copy and pasted from my question. Thank you for your help :)
OK, thanks
 
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