# Does the iron melt? Radiative-heat transfer

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:

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

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

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

• 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? – Allure Jul 15 at 23:31
• If the solid can't reach its melting point there is no melting, the latent heat of fusion should not useful – Ugo Mela Jul 16 at 5:54
• However the original, molten metal would have to freeze to produce the energy that melts the other piece, so it's still necessary? – Allure Jul 16 at 6:15
• 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 – Ugo Mela Jul 16 at 8:36
• In that case one doesn't need to calculate - the system will eventually reach thermal equilibrium. – Allure Jul 16 at 8:52