Temperature difference by Dimensional analysis.

In summary, the conversation discusses using dimensional analysis to solve a problem involving the heat generated from the decay of uranium in water. The individual suggests using dependent variables such as volume, heat, radiation, and heat capacity, and sets up equations based on their dimensions. However, they are not convinced by the resulting equation and question the accuracy of their dependent variables. Another person suggests including parameters such as thermal conductivity of uranium and temperature of the water in the analysis.
  • #1
pogs
15
0
Homework Statement
A block of uranium is kept cool in water. How does the difference between the temperature in the centre of the block and the temperature in the water depend on the size of the block?
Relevant Equations
n = n_0*e^(-t/τ) : Radioactive Decay
Attempt at solution:

I wanted to try and solve this with dimensional analysis. I reasoned that I would chose the following dependent variables:
- [V] : Volume ( of the block)
- [Q] : Heat ( the radioactive decay would cause some heating of the water)
- [R]: Radiation
- [Cv]: Heat capacity

I'm thinking the amount of radiation is a factor, it causes the heating and thus the bigger the block the more the radiation, the more heating. Also the heat capacity may play a part.

Taking the dimensions of these I get the following
[V] = L^3
[Q] = M L^2 T ^-2
[R] = L^2 M T^-1
[Cv] = L^2 M K^-1 T

Now we want a temperature so we can set up an equation

Temp(K) = [V]^α * [Q]^β * [R]^γ *[Cv] ^δ = (L^3)^α*(M L^2 T ^-2)^β*(L^2 M T^-1)^γ * (L^2 M K^-1 T)^δ

Taking the dimensions we get

L = 0 = 3α +2β +2γ +2δ
T = 0 = -2β -γ +δ
M = 0 = β +γ +δ
K = 1 = -δ

But when I solve these equations I get
α = 0
β = 2
γ = -1
δ = -1

giving me an equation that says

Q^2/(R*C_v)

Which I'm not to convinced by... Also V disappears as alpha is zero. I guess my dependent variables are wrong somehow. Any ideas/tips?
 
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  • #2
pogs said:
I wanted to try and solve this with dimensional analysis
Not convinced that that is possible.

How about an approach along the line
Uranium decays. Heat generated ##\propto## ...​
Heat exchange with surrounding water ##\propto## ...​

?

##\ ##
 
  • Like
Likes DaveE
  • #3
Other parameters you should include are thermal conductivity of the uranium and temperature of the water.
 

1. What is dimensional analysis?

Dimensional analysis is a mathematical technique used to analyze and understand the relationships between physical quantities. It involves examining the dimensions (such as length, mass, and time) of different variables and using them to derive equations and solve problems.

2. How is temperature difference calculated using dimensional analysis?

Temperature difference can be calculated by comparing the dimensions of temperature in two different units. For example, if we want to convert from degrees Celsius to degrees Fahrenheit, we can use the equation ΔT = (9/5)Δt, where ΔT is the temperature difference and Δt is the change in temperature in Celsius.

3. What is the importance of dimensional analysis in scientific research?

Dimensional analysis is important in scientific research because it allows for the simplification and understanding of complex relationships between physical quantities. It also helps to identify and correct errors in equations and measurements, and can be used to make predictions and solve problems in various fields of science.

4. Can dimensional analysis be used for non-physical quantities?

Yes, dimensional analysis can be used for non-physical quantities such as economic variables, chemical reactions, and even social phenomena. It is a versatile tool that can be applied to any situation involving quantities with different dimensions.

5. What are the limitations of dimensional analysis?

While dimensional analysis is a powerful tool, it has some limitations. It assumes that all relevant variables can be expressed in terms of fundamental dimensions, which may not always be the case. It also does not take into account the numerical values of variables, only their dimensions, so it cannot account for factors such as coefficients or constants in equations.

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