Calculating the Heat Capacity of Diamond

In summary, heat capacity is a physical property of a substance that stores energy. Diamond is highly conductive so the heat capacity is multiplied by the conductivity to get the total amount of heat that the diamond can store.
  • #1
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.
 
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  • #2
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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  • #3
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?
 
  • #4
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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  • #5
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.
 
  • #6
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
 
  • #7
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.
 
  • #8
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
 

FAQ: Calculating the Heat Capacity of Diamond

1. What is the heat capacity of diamond?

The heat capacity of diamond is approximately 6.01 J/g·K at room temperature.

2. How is the heat capacity of diamond calculated?

The heat capacity of diamond can be calculated using the formula: C = Q/mΔT, where C is the heat capacity, Q is the amount of heat energy absorbed or released, m is the mass of the diamond, and ΔT is the change in temperature.

3. What factors affect the heat capacity of diamond?

The heat capacity of diamond can be affected by factors such as temperature, pressure, and impurities. Changes in these factors can alter the crystal structure of diamond and thus affect its heat capacity.

4. How does the heat capacity of diamond compare to other materials?

Diamond has a relatively high heat capacity compared to most other materials. This is due to its strong covalent bonds and rigid crystal structure, which allows it to absorb and store a large amount of heat energy.

5. Why is calculating the heat capacity of diamond important?

Calculating the heat capacity of diamond is important for understanding its thermal properties and potential applications. It can also provide valuable information for industries that use diamond in high-temperature environments, such as in electronics and cutting tools.

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