Calculating Heat: How to Apply Calculus for Accurate Results

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SUMMARY

The discussion focuses on applying calculus to accurately calculate the heat required to raise the temperature of one mole of diamond from 20K to 100K using Debye's Law, which states that specific heat capacity (C) is proportional to (t/θ)^3. For diamond, the characteristic temperature θ is 1860K. The user initially attempted a discrete summation of heat increments but realized that integration is necessary for a precise calculation. The correct approach involves integrating the expression [A/(θ)^3] T^4/4 between the limits of 20K and 100K.

PREREQUISITES
  • Understanding of Debye's Law for specific heat capacity
  • Basic calculus concepts, particularly integration
  • Familiarity with thermodynamic principles
  • Knowledge of specific heat calculations
NEXT STEPS
  • Study the integration of functions to apply in thermodynamic calculations
  • Explore advanced applications of Debye's Law in different materials
  • Learn about specific heat capacity variations with temperature
  • Investigate the implications of low-temperature physics on material properties
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Students preparing for physics or materials science exams, educators teaching thermodynamics, and researchers interested in low-temperature material behavior.

s7b
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I'm doing some sample problems to prepare for midterm and am stuck on this:

The question talked about many materials at low temperatures obeying Debye's Law C=A(t/θ)^3

it said that for a diamond θ is 1860K and asked to evaluate the specific heats at 20K and 100K.

For this I just used that formula given. The part I'm having trouble with is how much heat is required to heat one mole of diamond between 20K and 100K.

I know that
To heat the diamond from 20 to 21 K, you need:

0.0024 J/molK

from 21 to 22 K, you probably need a little more

0.0026 J/molK ( more or less)

and so on until you heat it from 99 to 100 K where you need:

0.301 J/molK

so you need to add up

0.0024 + 0.0026 + ... +..... + 0.301 to get to the final answer - it should probably look like:

(0.0024 + 0.301) / 2 x (100 - 20 K) = 12.136 J/mol

But this is not the correct answer. How can I apply calculus in order to obtain a more correct answer?
 
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Well, I have not met that Debye dude, but I think that you need to simply integrate and evaluate the expression at the two limits

your integral would result in something like this: [A/(θ)^3] T^4/4

evaluate that expression at 100 and 20 and subtract such numbers.

does this help? maybe?
 
This makes more sense then my attempt. Thanks :)
 

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