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Piece of iron put into container with ice
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[QUOTE="Lukasz Madry, post: 6004114, member: 646364"] [h2]Homework Statement [/h2] We put 1kg iron of temperature 100 Celsius into container with 1kg of ice, temperature 0 Celsius. What is state of the system after reaching equilibrium? Calculate change of entropy. Coefficient of melting of ice (c_L) is 330 kJ/kg, coefficient of heat transfer of iron (c_I) is 450 J/(kg*K).[h2]Homework Equations[/h2] I don't really know. Heat balance, for sure. Entropy equation - Q = ∫ T dS. S = c ln(T_f/T_i) But I'm sure I'm missing something crucial. [h2]The Attempt at a Solution[/h2] I tried to compute heat balance: C_I ( 100 - T_f) = c_L but this does not work, obviously, since leads to T_f = 100 - c_L/c_I = 100 - 733 which is far below zero temperature. Another attempt - assume that only part of ice was melted. Let k = c_I 100 / c_L be coefficient that describes how much ice was melted into water. In this case, entropy would be: S_{iron} = c_I ln(273/373) S_{ice} = k c_L/273 after substition we can find that it's sum is greater than zero, which is expected. Is this close to being correct? [/QUOTE]
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Piece of iron put into container with ice
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