# B Calculating heat from decay of a radioactive isotope

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1. Feb 24, 2017

### COWilliam

How would one go about determining the amount of heat generated by the decay of a radioactive particle, such as Cesium 137, Polonium 210, or Strontium 90? How would you determine how much of the radioactive material would be needed to heat, say, a cup of water to a certain temperature, taking into account thermal resistance and volume of the material?

2. Feb 24, 2017

### haruspex

You would need to know the mass loss in the decay.

3. Feb 25, 2017

### COWilliam

Wouldn't that be almost insignificant? I mean, even a fast decaying particle such as Polonium, which has a pretty short half life of a hundred and thirty-something days is not going to have a major effect on the amount of heat generated in a short period that it would take to heat a cup of water.

4. Feb 25, 2017

### haruspex

How else will heat be generated?

5. Feb 25, 2017

### Staff: Mentor

Look up the energy released per decay, calculate how many decays happen per time and mass, multiply both to get the power density. Find the power you want to get the amount of material you need.

It is impractical.

6. Feb 25, 2017

### COWilliam

Lets take Polonium 210, which has a half life of 138 days, has a (TBq/g) activity of 166, produces 140 watts/g, decays into Alpha radiation, and has 5.3 Million electron volts of alpha energy. If we put that into a gallon of water, insulated in say a quarter inch of steel, how much Polonium would be needed to raise the temperature to 37oC? What formulas would be used?

7. Feb 25, 2017

### haruspex

That would not be very good thermal insulation. If you have a slow source of heat then you need pretty good thermal insulation.

8. Feb 25, 2017

### COWilliam

OK, what would you suggest I use to insulate it?

9. Feb 25, 2017

### COWilliam

I suppose the most effective insulation would be a vacuum, so putting it inside of a vacuum chamber would be pretty good for such a small heat source...

10. Feb 25, 2017

### Astronuc

Staff Emeritus
Follow mfb's advice, and determine the energy needed to increase temperature (ΔT) of a given mass of water, which is pretty straight-forward. Then use the energy per decay to determine the number of decays/atoms necessary to provide that energy. If one uses power, W/g, then one has to integrate over time to establish the number of atoms decaying during some period. Simply assume adiabatic conditions, and don't make the problem hard than it is.

If one needs a power, then simply use the W/g.

11. Feb 25, 2017

### haruspex

Make that a vacuum flask, which has a silvered surface to minimise radiation. You could look up the thermal characteristics of those.

12. Feb 25, 2017

### COWilliam

I would prefer to not assume that I would have a perfect adiabatic system, but since most commercially available VIPs reduce the amount of transferred heat to near inconsequential values, I suppose I can ignore this for the time being. Thanks.

13. Feb 25, 2017

### Astronuc

Staff Emeritus
The alternative is to use a well-insulated system and determine the rate at which the system is losing heat to the environment. Is one assuming a steady-state situation, or a transient (time-dependent) situation? If there is heat loss from the system, which may be a strong or weak function of temperature, then one would have to calculate the amount of energy necessary to heat the water and the heat lost to the environment.

14. Feb 25, 2017

### Staff: Mentor

Vacuum flasks can keep water hot for hours. Unless you want to wait hours to heat something up, you can find insulation that is good enough to ignore heat losses.

15. Feb 25, 2017

### COWilliam

OK, so if I had 1 gram of Polonium 210, which produces 140 Joule/s, and it takes about 4 Joules to raise the temperature of 1 gram of water by 1oC, to raise the temperature of 1 kilogram of water to from 23oC to 30oC would be 28000 Joules, that would be delivered over about 200 second? Is all the energy being produced going to be absorbed as heat?

16. Feb 25, 2017

### haruspex

Pretty much.

17. Feb 25, 2017

### COWilliam

OK, fantastic. That was all very helpful guys.

18. Feb 26, 2017

### snorkack

... in case of alpha decay.

19. Feb 26, 2017

### haruspex

Is it not true of any decay in which the products are contained, so unable to carry away the energy?

20. Feb 26, 2017

### COWilliam

I would imagine so, however radiation such as beta and gamma tends to pass through most materials without reacting to them. Alpha particles are absorbed extremely easily and quickly, even a sheet of paper can block them.

21. Feb 28, 2017

### snorkack

Also a lot of beta decay energy goes to neutrinos.