Whats the materials specific heat capacity?

Click For Summary
SUMMARY

The discussion centers on calculating the specific heat capacity of a material using the formula Q = mcΔT. The initial calculation presented by the user Dx was 1.35 kcal/kg°C, which was later corrected to 0.34 kcal/kg°C after re-evaluating the temperature conversion from Fahrenheit to Celsius. The participants confirmed the correct approach and highlighted the importance of accurate unit conversion in thermodynamic calculations.

PREREQUISITES
  • Understanding of the specific heat capacity formula Q = mcΔT
  • Knowledge of temperature conversion between Fahrenheit and Celsius
  • Basic principles of thermodynamics
  • Familiarity with scientific notation and unit analysis
NEXT STEPS
  • Study the principles of thermodynamics and heat transfer
  • Learn detailed methods for temperature conversion, specifically between Fahrenheit and Celsius
  • Explore examples of specific heat capacity calculations in different materials
  • Investigate common mistakes in unit conversions and how to avoid them
USEFUL FOR

Students in physics or chemistry, educators teaching thermodynamics, and professionals involved in materials science or engineering who require accurate calculations of specific heat capacities.

Dx
Hello,

150kcal of heat raises the temp of 2.0kg of material by 400 degrees F. Whats the materials specific heat capacity?

I cam up with 1.35 kcal/kg degrees C. is this correct?

Thanks!
Dx :wink:
 
Last edited by a moderator:
Physics news on Phys.org
Dx,

That's not what I get. Can you show me what you did?

Thanks,
 
I used some substitution

Originally posted by Tom
Dx,

That's not what I get. Can you show me what you did?

Thanks,

No problem, Tom! I meant to say .34 kcal/fg oC was my answer sorry that's another problem i am working on but i got it wrong anyways, lol.

I used the formula [del]Q = mc [del]T and substituted.
c = 150kcal / (2kg)* (190oC) =
.34 kcal

i figured it was .34 since when i converted from F to C i rounded the number so then I usd scientific notation to conclude to this answer.
I am sorry but I have reworked this problem many times erasing and adding to it so excuse me if its not exact but you can see I am really trying here.
Thanks!
Dx
 
OK, that's what I got too.

I was thinking you had made a mistake converting ΔTF to ΔTC (because you have to multiply by 5/9), but then I reworked the problem by making that mistake, and I still did not get your answer.

Oh well, that's cleared up now.
 
TY Tom

Just wanted to say thanks!
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
830
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K