Specific heat capacity of a solid material

In summary: I'll keep trying to get ahold of the software so that I can trouble shoot it myself.OK, thanks for getting back to me. I'll keep trying to get ahold of the software so that I can trouble shoot it myself.
  • #1
Aleisha
25
0

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
 
Physics news on Phys.org
  • #2
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.)
 
  • #3
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.
 
  • #4
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?
 
  • #5
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?
 
  • #6
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%.
 
  • Like
Likes Aleisha
  • #7
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 :)
 
  • #8
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.
 
  • #9
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 :)
 
  • #10
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
 

What is specific heat capacity?

Specific heat capacity is a measure of the amount of heat required to raise the temperature of a material by one degree Celsius per unit mass. It is a physical property that is specific to each material and is often denoted by the symbol "C".

How is specific heat capacity measured?

Specific heat capacity is usually measured by performing experiments in which a known amount of heat is added to a sample of the material, and the resulting change in temperature is recorded. The specific heat capacity can then be calculated using the equation Q = m*C*ΔT, where Q is the heat added, m is the mass of the material, and ΔT is the change in temperature.

What factors affect the specific heat capacity of a material?

The specific heat capacity of a material can be affected by several factors, including the mass and composition of the material, its phase (solid, liquid, or gas), and the temperature at which it is measured. It can also vary slightly depending on the experimental conditions and the accuracy of the measurement.

Why is specific heat capacity important?

The specific heat capacity of a material is important because it is a crucial factor in determining how much energy is required to change its temperature. This plays a significant role in many practical applications, such as designing heating and cooling systems, calculating the energy requirements for industrial processes, and understanding the behavior of materials in extreme conditions.

How does the specific heat capacity of a material differ from its heat capacity?

The specific heat capacity of a material is a property that is specific to a particular material and is measured per unit mass. On the other hand, the heat capacity of a material is a measure of the amount of heat required to raise the temperature of the entire object, regardless of its mass. The specific heat capacity is therefore more useful for comparing the heat storage capacities of different materials, while the heat capacity is more relevant for determining the overall energy requirements of a system.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
974
  • Introductory Physics Homework Help
Replies
2
Views
793
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
760
  • Introductory Physics Homework Help
Replies
3
Views
3K
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
747
Back
Top