Calculating Specific Heat Capacity

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Homework Help Overview

The discussion revolves around calculating the specific heat capacity of a solid material that is heated from a lower to a higher temperature using a specified amount of energy, factoring in the efficiency of the heating process.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of specific heat capacity using the formula Q = (m)(c)(change in T) and express confusion regarding the application of efficiency in their calculations. There are attempts to clarify the correct use of efficiency and how it affects the energy input.

Discussion Status

Some participants have provided calculations and expressed uncertainty about the correctness of their answers, particularly regarding significant figures and the units of the final result. There is a recognition that the calculations appear correct, but the input format may be causing issues.

Contextual Notes

Participants mention specific requirements for presenting the answer, including units and significant figures, which may be influencing their perceived correctness of the solution.

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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