If the warhead of the Oreshnik is very, very hot, how much heat might it hold? Since it most likely consists of different components, we would need to know the specific heat capacity of each and how warm they are. While some may reach a temperature of 3900, that is most likely not all. It is complicated to evaluate the total amount of stored heat in the missile on impact, but worth trying.
Using this calculator I found some values for the heat capacity of common materials:
Ice at -10 C (solid): 2.05 J/g*K
Water at 25 C: 4.1813 J/g*K
Water at 100 C (gas): 2.05 J/g*K
Animal tissue: 3.5 J/g*K (A body contains mostly water, though for humans it depends on sex and constitution, women are on average less watery, and so are people with a significant fatty tissue percentage. )
The values for metals are compared to the heat capacity of water low, mostly:
Tungsten: 0.134 J/g*K
Uranium: 0.116 J/g*K
Armour penetrating ammunition can be made from heavy metals similar to the two above.
Bismuth (solid) 0.133 J/g*K
Silver: 0.233 J/g*K
Copper (solid): 0.385 J/g*K
Steel (solid): 0.466 J/g*K
Titanium (solid) 0523 J/g*K
Aluminium (solid): 0.897 J/g*K
Sodium (solid): 1.23 J/g*K
Lithium (solid): 3.58 J/g*K
Lithium at 181 C (liquid): 4.379 J/g*K
Then there is are the heat capacity of non-metals:
Glass, silica (solid): 0.84 J/g*K
Diamond: 0.509 J/g*K
Brick (solid): 0.84 J/g*K
Air (at sea level): 1.00 J/g*K (The mass of 1 liter of air at 15 C is listed as 1.255 g. Air consists mainly of nitrogen and oxygen and the values of both are similar, but with the first being slightly higher than the second)
If ceramic materials are used in the construction of the missiles, it turns out that for some ceramic metals, the heat capacity increases sharply with the temperature, which even to begin with is higher than for many metals. For an idea, compare that of silica with that of tungsten or iron.
The abstract of this paper Thermal Conductivity and Specific Heat Capacity of Different Compositions of Yttria Stabilized Zirconia-Nickel Mixtures reads:
Using the data to calculate the accumulated heat energy in a 100 kg warhead
I am not sure about how to make a good model, using the data above, to evaluate the accumulated heat energy in the warhead, because there is so much we don't know, but what is lost from trying? So here we go:
Assume first that the warhead has a mass of 100 kg (100,000 g) and an average heat capacity of 1 J/g*K which is higher than one would expect for many metals and silica, but still includes the possibility of a sharp increased heat capacity of some ceramic metals.
Next assume that the average temperature of the warhead is 1000 C above the temperature of the environment, this is more than most electronics can sustain, but if it is well insulated form more hot parts, who knows?
Inserting these numbers using the formula for heat, we can estimate the stored energy,
Q = mc ΔT where, Q= heat, m = mass of the body, c = specific heat, and ΔT = temperature difference.
This gives inserting the previous values,
Q=100,000g * 1J/g*K * 1000K
The above gives
Q = 100,000,000 J which can also be written as 100 *10^6 J or 100 MJ
If we add 100 MJ of stored heat energy to the estimated kinetic energy of 200 MJ, as estimated in the earlier post, there is a total of 300 MJ. It is enough to evaporate all the water in a body, about 150 MJ according to this calculation of the energy assuming the initial temperature is 37 degrees.
The Specific Heat Capacity of water is approximately 4.185 J/g°C
From 37 to 100 degrees C, there are 63 degrees, one gram of water at 37 degrees heated to 100 would require 63*4.185 J/g°C=264 J
Add to this the energy needed to evaporate the water 2260 J
From 37 degrees Celsius to vapor requires 2524 J For one kg we need, 2.524 MJ
Assuming the body of a man consists of 60 % water and the person weighs a 100 kg then there would be 60 kg to evaporate. This would require 60* 2.524 MJ or roughly 150 MJ, and half for a person of 50 kg (30 kg of water).
However it is not sufficient for an outright cremation. allegedly 1,400 -2,500 MJ according to this Quora answer
This ends all the calculations and estimates. From future observations it will be possible to make adjustments, but already now there is a rough idea.
Reflection
After the process of working on the problem, I woke up wondering if there actually are people in power, in the military industrial complex, and among active duty officers who would press for provoking Russia just to make sure it shows more cards, including using the weapons it has, some of which they are unable to create themselves, but would like to copy and make too? No matter the underlying motives, in the process of pressing for escalation they cause the deaths of hundreds of thousands fo people, and risk the lives of millions and billions. They are completely irresponsible.
Using this calculator I found some values for the heat capacity of common materials:
Ice at -10 C (solid): 2.05 J/g*K
Water at 25 C: 4.1813 J/g*K
Water at 100 C (gas): 2.05 J/g*K
Animal tissue: 3.5 J/g*K (A body contains mostly water, though for humans it depends on sex and constitution, women are on average less watery, and so are people with a significant fatty tissue percentage. )
The values for metals are compared to the heat capacity of water low, mostly:
Tungsten: 0.134 J/g*K
Uranium: 0.116 J/g*K
Armour penetrating ammunition can be made from heavy metals similar to the two above.
Lead (solid) 0.127 J/g*K
Bismuth (solid) 0.133 J/g*K
Silver: 0.233 J/g*K
Copper (solid): 0.385 J/g*K
Steel (solid): 0.466 J/g*K
Titanium (solid) 0523 J/g*K
Aluminium (solid): 0.897 J/g*K
Sodium (solid): 1.23 J/g*K
Lithium (solid): 3.58 J/g*K
Lithium at 181 C (liquid): 4.379 J/g*K
Then there is are the heat capacity of non-metals:
Glass, silica (solid): 0.84 J/g*K
Diamond: 0.509 J/g*K
Brick (solid): 0.84 J/g*K
Air (at sea level): 1.00 J/g*K (The mass of 1 liter of air at 15 C is listed as 1.255 g. Air consists mainly of nitrogen and oxygen and the values of both are similar, but with the first being slightly higher than the second)
If ceramic materials are used in the construction of the missiles, it turns out that for some ceramic metals, the heat capacity increases sharply with the temperature, which even to begin with is higher than for many metals. For an idea, compare that of silica with that of tungsten or iron.
The abstract of this paper Thermal Conductivity and Specific Heat Capacity of Different Compositions of Yttria Stabilized Zirconia-Nickel Mixtures reads:
If the thermal conductivity is low at high temperatures, a metal ceramic of the above type would also give off the heat slowly. An everyday example of differences in thermal conductivity between different materials is that we can hold a cup of warm tea more easily if the cup is made of ceramics than if it made of stainless steel, because steel conducts heat better than ceramics. Although on its own the heat conductivity is low of metal ceramics what happens if they are crushed and mixed with other components on impact?
Using the data to calculate the accumulated heat energy in a 100 kg warhead
I am not sure about how to make a good model, using the data above, to evaluate the accumulated heat energy in the warhead, because there is so much we don't know, but what is lost from trying? So here we go:
Assume first that the warhead has a mass of 100 kg (100,000 g) and an average heat capacity of 1 J/g*K which is higher than one would expect for many metals and silica, but still includes the possibility of a sharp increased heat capacity of some ceramic metals.
Next assume that the average temperature of the warhead is 1000 C above the temperature of the environment, this is more than most electronics can sustain, but if it is well insulated form more hot parts, who knows?
Inserting these numbers using the formula for heat, we can estimate the stored energy,
Q = mc ΔT where, Q= heat, m = mass of the body, c = specific heat, and ΔT = temperature difference.
This gives inserting the previous values,
Q=100,000g * 1J/g*K * 1000K
The above gives
Q = 100,000,000 J which can also be written as 100 *10^6 J or 100 MJ
If we add 100 MJ of stored heat energy to the estimated kinetic energy of 200 MJ, as estimated in the earlier post, there is a total of 300 MJ. It is enough to evaporate all the water in a body, about 150 MJ according to this calculation of the energy assuming the initial temperature is 37 degrees.
The Specific Heat Capacity of water is approximately 4.185 J/g°C
From 37 to 100 degrees C, there are 63 degrees, one gram of water at 37 degrees heated to 100 would require 63*4.185 J/g°C=264 J
Add to this the energy needed to evaporate the water 2260 J
From 37 degrees Celsius to vapor requires 2524 J For one kg we need, 2.524 MJ
Assuming the body of a man consists of 60 % water and the person weighs a 100 kg then there would be 60 kg to evaporate. This would require 60* 2.524 MJ or roughly 150 MJ, and half for a person of 50 kg (30 kg of water).
However it is not sufficient for an outright cremation. allegedly 1,400 -2,500 MJ according to this Quora answer
This ends all the calculations and estimates. From future observations it will be possible to make adjustments, but already now there is a rough idea.
Reflection
After the process of working on the problem, I woke up wondering if there actually are people in power, in the military industrial complex, and among active duty officers who would press for provoking Russia just to make sure it shows more cards, including using the weapons it has, some of which they are unable to create themselves, but would like to copy and make too? No matter the underlying motives, in the process of pressing for escalation they cause the deaths of hundreds of thousands fo people, and risk the lives of millions and billions. They are completely irresponsible.
