Question about black bodies, emissivity.

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In a scenario involving a diamond sphere with a radius of 1 cm and a magic heater producing 0.3 W indefinitely, the discussion centers on the implications of zero emissivity. With zero emissivity, the sphere does not emit or absorb heat, leading to the conclusion that heat cannot escape. As the heater continuously adds energy, the internal heat builds up without any loss, resulting in an infinite temperature over an infinite time. This phenomenon highlights the relationship between heat energy, heat capacity, and temperature, where finite heat capacity combined with infinite energy leads to infinite temperature. The discussion emphasizes the theoretical nature of this scenario in thermodynamics.
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Imagine a diamond sphere radius of 1 cm traveling in space far from any heat sources. Embedded in the center of the sphere is a magic heater that can produce 0.3 W for an indefinite amount of time.

If the emissivity of the surface is zero, what temperature does the sphere reach? Why?

The answer is infinite temperature. I don't understand this. If emissivity is zero, doesn't that mean the body doesn't emit OR absorb any heat? I would think that the answer is that it would not reach any temperature... Can somebody explain to me why it would be infinite temperature? Is it because the time factor?
 
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I'm guessing that, because the emissivity is zero, there is no way for heat energy to escape from the diamond through radiation. So heat keeps building up indefinitely at a rate of 0.3 J per second inside the diamond, reaching infinity after an infinite amount of time. If the heat capacity of the diamond is finite, then the temperature at that time is infinite, since, (temperature) = (heat energy) / (heat capacity)
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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