Calculating Heat: How to Apply Calculus for Accurate Results

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In summary, the conversation was about a question involving Debye's Law and the specific heats of diamond at different temperatures. The formula was used to calculate the heat required to heat one mole of diamond between 20K and 100K. However, the correct answer was not obtained and the use of calculus was suggested to get a more accurate answer. The person providing the summary has not heard of Debye's Law but suggests integrating and evaluating the expression at the two limits to get the correct answer.
  • #1
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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  • #2
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?
 
  • #3
This makes more sense then my attempt. Thanks :)
 

1. What is the relationship between calculus and heat calculations?

Calculus is a branch of mathematics that deals with the study of change and motion. Heat calculations involve the determination of the transfer of thermal energy between objects. Calculus allows for precise and accurate calculations of heat by using mathematical models to describe the rate of change of temperature over time.

2. How is calculus used to calculate heat?

Calculus is used to calculate heat by applying the principles of differentiation and integration to equations that describe the transfer of thermal energy. This allows for the determination of rates of change and total amounts of heat transferred in a given system.

3. What are the key concepts of calculus that are relevant to heat calculations?

The key concepts of calculus that are relevant to heat calculations are derivatives, integrals, and differential equations. Derivatives are used to determine rates of change, while integrals are used to determine total amounts. Differential equations describe the relationship between a function and its derivatives, providing a framework for heat calculations.

4. What types of problems can be solved using calculus for heat calculations?

Calculus can be used to solve a wide range of problems in heat calculations, including determining the rate of heat transfer in a system, finding the maximum or minimum temperature reached, and calculating the total amount of heat transferred over a certain period of time. It can also be used to model more complex systems, such as heat conduction through different materials.

5. How can I apply calculus to real-world heat calculation problems?

In order to apply calculus to real-world heat calculation problems, it is important to first understand the underlying principles and equations involved. This includes knowledge of heat transfer mechanisms, such as conduction, convection, and radiation, as well as the relevant mathematical models. From there, it is a matter of setting up and solving the appropriate differential equations to obtain accurate results.

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