This discussion focuses on calculating the heat generated from the decay of radioactive isotopes, specifically Cesium 137, Polonium 210, and Strontium 90. Key calculations involve determining the energy released per decay, the number of decays per time and mass, and the necessary amount of radioactive material to heat a specific volume of water. For Polonium 210, which has a half-life of 138 days and produces 140 watts per gram, the discussion emphasizes the importance of thermal insulation, suggesting a vacuum flask for effective heat retention. Participants agree on the need to account for energy loss to the environment and the significance of using precise formulas to achieve accurate results.
PREREQUISITESThis discussion is beneficial for physicists, nuclear engineers, and anyone involved in thermal management or radioactive material handling, particularly in applications requiring precise heat calculations from radioactive decay.
You would need to know the mass loss in the decay.COWilliam said: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?
How else will heat be generated?COWilliam said: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.
That would not be very good thermal insulation. If you have a slow source of heat then you need pretty good thermal insulation.COWilliam said:insulated in say a quarter inch of steel
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.COWilliam said: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?
Make that a vacuum flask, which has a silvered surface to minimise radiation. You could look up the thermal characteristics of those.COWilliam said:I suppose the most effective insulation would be a vacuum
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.Astronuc said: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.
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.COWilliam said: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.
Pretty much.COWilliam said:Is all the energy being produced going to be absorbed as heat?
... in case of alpha decay.haruspex said:Pretty much.
Is it not true of any decay in which the products are contained, so unable to carry away the energy?snorkack said:... in case of alpha decay.
Also a lot of beta decay energy goes to neutrinos.COWilliam said:I would imagine so, however radiation such as beta and gamma tends to pass through most materials without reacting to them.