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A molten metal (iron type A) is in vacuum (no air) at a constant temperature ($T_m$) equal to 1500 C (equal to its melting temperature) with emissivity $\epsilon_m$.

At a distance of 10 cm there is a block of solid iron (type A) at temperature ($T_s$) 1100 C with emissivity $\epsilon_s$.

I know that at the surface of the solid (facing the liquid), for the heat balance:

$Q = G - \rho*G + \epsilon_s*E_b$

where:

  • $G = \epsilon_m*\sigma * T_m^4$, $\sigma$ = Stefan-Boltzmann constant
  • $\rho$ refectivity of the surface of solid metal = $1 - \epsilon_s$
  • $E_b = \sigma * T_s^4$

so:

$Q = \epsilon_s * G + \epsilon_s*E_b = \epsilon_s * \sigma * ( \epsilon_m * T_m^4 - T_s^4) = \epsilon_s * \sigma * ( ( \sqrt[4]{\epsilon_m} * T_m)^4 - T_s^4) $

so the two surface can't reach the same temperature: the solid metal never melts because the maximum temperature that it can reach is $\sqrt[4]{\epsilon_m} * T_m$.

1. Is it correct?
2. If yes, Is it a real scenario?

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  • $\begingroup$ Wouldn't you need to know the latent heat of fusion (i.e. how much energy is required to melt iron type A) to solve this? $\endgroup$ – Allure Jul 15 at 23:31
  • $\begingroup$ If the solid can't reach its melting point there is no melting, the latent heat of fusion should not useful $\endgroup$ – Ugo Mela Jul 16 at 5:54
  • $\begingroup$ However the original, molten metal would have to freeze to produce the energy that melts the other piece, so it's still necessary? $\endgroup$ – Allure Jul 16 at 6:15
  • $\begingroup$ Maybe I didn't write it down but you can assume that the molten metal is always liquid: there is a source of heat (or similar) that heats it to maintain that temperature $\endgroup$ – Ugo Mela Jul 16 at 8:36
  • $\begingroup$ In that case one doesn't need to calculate - the system will eventually reach thermal equilibrium. $\endgroup$ – Allure Jul 16 at 8:52

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