Calculating the Heat Capacity of Diamond

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SUMMARY

The heat capacity of diamond is quantified as 6.57 Joules/mole per degree Kelvin, indicating its ability to store energy. The relevant formulas for this discussion include q=Cp (ΔT/Δt) for heat input and q/A=k (ΔT/Δx) for heat conduction, with diamond's thermal conductivity measured at 895 W/m-K. The discussion centers on determining the temperature increase of diamond subjected to a bombardment of 3.7x10^10 beta particles per second, each with an energy of 182 keV, while considering the effects of insulation and heat loss. The consensus is that without additional information on heat transfer coefficients and the insulation's K value, the problem remains indeterminate.

PREREQUISITES
  • Understanding of heat capacity and its calculation
  • Familiarity with thermal conductivity and its measurement
  • Knowledge of beta decay and energy calculations
  • Basic principles of heat transfer and insulation properties
NEXT STEPS
  • Research the thermal conductivity of various insulating materials and their K values
  • Explore the principles of steady-state heat transfer in solids
  • Learn about the effects of radiation on temperature in insulated systems
  • Investigate the mathematical modeling of heat transfer in materials subjected to particle bombardment
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Students in physics or materials science, engineers working with thermal management, and researchers studying the thermal properties of diamond and similar materials.

MDM
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Homework Statement



Heat capacity is the ability of the material to store energy internally. If I completely insulated diamond and I put heat into it, It would have the ability to store 6.57 (Joules/mole) per degree Kelvin. Use this formula q=Cp (ΔT/ Δt) where q is heat in Watts, ΔT is differential temperature and Δt is differential time (obviously, Cp is heat capacity) and Watts = .5 mW.

Diamond has a very high conductivity. q/A=k (ΔT/ Δx) where A is area, k is conductivity of diamond at 895 W/m-K, and Δx is the differential distance.

Let's say there is a source that produces 3.7x10^10 beta particles per second. A beta from this decay has an average energy of 182 keV. With that, over time, how hot will the outside get?

Is this solvable, if it is not what information do you need and if it is how do you solve it?

Is this definitely a chemistry question?

Homework Equations



q=Cp (ΔT/ Δt)

q/A=k (ΔT/ Δx)

The Attempt at a Solution



Try to solve as a differential equation. Did not work. Could not find any relevant guides that showed how to solve this particular heat capacity formula.
 
Last edited:
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MDM said:

Homework Statement



Heat capacity is the ability of the material to store energy internally. If I completely insulated diamond and I put heat into it, It would have the ability to store 6.57 (Joules/mole) per degree Kelvin. Use this formula q=Cp (ΔT/ Δt) where q is heat in Watts, ΔT is differential temperature and Δt is differential time (obviously, Cp is heat capacity) and Watts = .5 mW.

Diamond has a very high conductivity. q/A=k (ΔT/ Δx) where A is area, k is conductivity of diamond at 895 W/m-K, and Δx is the differential distance.

Is this solvable, if it is not what information do you need and if it is how do you solve it?

Is this definitely a chemistry question?

Homework Equations



q=Cp (ΔT/ Δt)

q/A=k (ΔT/ Δx)

The Attempt at a Solution



Try to solve as a differential equation. Did not work. Could not find any relevant guides that showed how to solve this particular heat capacity formula.
The heat capacity of a substance is a physical property, like density.

If q is measured in watts, then it's not a heat input but a heat input rate, or q-dot, since watts are derived units, such that 1 watt = 1 Joule / s.

It's not entirely clear what you are looking for here. Are you trying to determine the change in temperature versus time of a diamond with a heat input rate of 0.5 watt?
 
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SteamKing said:
The heat capacity of a substance is a physical property, like density.

If q is measured in watts, then it's not a heat input but a heat input rate, or q-dot, since watts are derived units, such that 1 watt = 1 Joule / s.

It's not entirely clear what you are looking for here. Are you trying to determine the change in temperature versus time of a diamond with a heat input rate of 0.5 watt?

Left out something important here. Let's say there is a source that produces 3.7x10^10 beta particles per second. A beta from this decay has an average energy of 182 keV. With that, over time, how hot will the outside get?
 
Are you looking for the surface temperature when the system reaches steady state so that the temperature is no longer changing? Are all surfaces of the diamond being bombarded, or just one surface? Do you know how many particles hit the surface per second, and do you know the area of the part of the surface receiving the bombardment? Do you have any estimate of what the temperature might be so that we know whether to include radiant heat transfer?
 
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Great questions! Only one side is being bombarded (it's encapsulated). The source of the beta particles is 90Sr. With 90Sr you will get a beta from its decay with an average energy of 182 keV and a beta from 90Yr with an average energy of 760 keV. We're looking for the steady state if there is one. 20mm/35mm = 700mm area

No estimate of temperature. 3.7x10^10 beta particles per second.
 
OK. You've given the area that's being bombarded. How about the remainder of the surface area. How much is that? Since the diamond is highly conductive, we are going to assume (to start with) that the entire diamond is at a uniform temperature. But, to find out what that temperature is, we need to estimate the rate of heat loss, so that we can match it to the rate of heat gain from bombardment. The diamond surface is going to be much hotter than the bulk room temperature air. The rate of heat loss is proportional to the temperature difference between the surface and the room air. The constant of proportionality is the area for heat loss times the so-called "heat transfer coefficient." We are not going to know the heat transfer coefficient exactly, but we are going to use a range of typical measured values from the literature. This will give us the possible range of diamond temperatures.

Later, we can consider the effect of heat conduction in the diamond. At steady state, the heat capacity of the diamond is not a significant parameter (unless we are interested in estimating how long it will take for the diamond to heat up).

Chet
 
The initial statement includes "If I completely insulated diamond ".

Since we do not have a heat transfer value for the insulation, the diamond should be assumed to be perfectly insulated; and, in that case, it would appear to me that this is a total heat input problem and the answer is indeterminate without knowing the total bombardment time.

If the the insulation is not intended to be perfect; then, we have a heat transfer problem; and, the required missing information is the insulation's K value.

Does anyone else see it this way; or am I completely off base here.
 
JBA said:
The initial statement includes "If I completely insulated diamond ".

Since we do not have a heat transfer value for the insulation, the diamond should be assumed to be perfectly insulated; and, in that case, it would appear to me that this is a total heat input problem and the answer is indeterminate without knowing the total bombardment time.

If the the insulation is not intended to be perfect; then, we have a heat transfer problem; and, the required missing information is the insulation's K value.

Does anyone else see it this way; or am I completely off base here.
You're right. I didn't read the OP carefully enough. Sorry. The implications of correctly reading the description are that the diamond will not reach steady state, and its temperature will continue to rise as a function of time (pretty linearly with time), unless heat can be radiated from the bombardment surface.

Thanks for spotting my mistake.

Chet
 

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