Calculating Specific Heat Capacity

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SUMMARY

The discussion focuses on calculating the specific heat capacity of a solid material weighing 4.96 kg, heated from 16.7°C to 234°C using 725 kJ of energy with an efficiency of 0.342. The correct formula for specific heat capacity is c = Q/(m * ΔT), where Q is the effective energy input. The participant calculated Q as 247950 J and derived a specific heat capacity of 230.05 J/kg/K, which was later rounded to 230 J/kg/K. Despite following the calculations correctly, the participant received feedback indicating the answer was still incorrect, raising questions about significant figures and reporting units.

PREREQUISITES
  • Understanding of specific heat capacity and its formula
  • Basic knowledge of energy efficiency calculations
  • Familiarity with significant figures in scientific reporting
  • Ability to perform unit conversions (e.g., J to kJ)
NEXT STEPS
  • Review the principles of energy efficiency in thermal processes
  • Learn about significant figures and their importance in scientific calculations
  • Explore unit conversion techniques, specifically between Joules and kilojoules
  • Investigate common pitfalls in specific heat capacity calculations
USEFUL FOR

Students in physics or chemistry courses, educators teaching thermodynamics, and anyone involved in material science or thermal energy calculations.

struggtofunc
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Homework Statement


A 4.96 kg piece of solid material is heated from 16.7oC to 234oC (3 s.f.) using 725 kJ of energy (3 s.f.).

Assuming an efficiency of 0.342 for the heating process, and that the material does not melt, calculate the specific heat capacity of the material.

m = 4.96 kg
change in T = 217.3 celsius
energy = 725000 J
efficiency = 0.342

Homework Equations


Q = (m)(c)(change in T)

The Attempt at a Solution


c = Q/(m*change in T)

I have calculated Q using .342*725000 = 247950.

I then plugged in these values into the equation and am still getting the wrong answer.

I've also tried dividing 725000/.342 and plugging those values in but it's still incorrect.

What should I be doing in regards to the efficiency?

Thanks!
 
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struggtofunc said:

Homework Statement


A 4.96 kg piece of solid material is heated from 16.7oC to 234oC (3 s.f.) using 725 kJ of energy (3 s.f.).

Assuming an efficiency of 0.342 for the heating process, and that the material does not melt, calculate the specific heat capacity of the material.

m = 4.96 kg
change in T = 217.3 celsius
energy = 725000 J
efficiency = 0.342

Homework Equations


Q = (m)(c)(change in T)

The Attempt at a Solution


c = Q/(m*change in T)

I have calculated Q using .342*725000 = 247950.

I then plugged in these values into the equation and am still getting the wrong answer.

I've also tried dividing 725000/.342 and plugging those values in but it's still incorrect.

What should I be doing in regards to the efficiency?

Thanks!
Please show the details of your calculation of c.
 
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c = 247950/(4.96*217.3) = 230. 05 J/kg/K
 
struggtofunc said:
c = 247950/(4.96*217.3) = 230. 05 J/kg/K
This looks OK, but isn't it usually reported as kJ/kg/K? Also, shouldn't the significant figures be less?
 
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Hi chestermiller,

Yes, I input the answer as 230 J/kg/K because it was what we were told to present it in. It's still showing as incorrect though.
 
struggtofunc said:
Hi chestermiller,

Yes, I input the answer as 230 J/kg/K because it was what we were told to present it in. It's still showing as incorrect though.
I don't know what else to suggest. This answer looks correct.
 
No worries. Thank you very much for your help!
 

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