# Piece of iron put into container with ice

## Homework Statement

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).

## Homework Equations

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.

## The Attempt at a Solution

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?