Suppose mixture of 500g of water and 100g of ice that both are in thermic equilibrium. We then add 200g of steam which has a temperature of 100°C. Calculate the equilibrium temperature and the composition of the final mixture.

Used formula's:

$Q = mL$ where Q = heat, m = mass, L = latent heat

$Q = mc\Delta t$ where m=mass, c=specific heat, $\Delta t$ = change in temperature


  • $m_{\text{water}} = 500g$

  • $m_{\text{ice}} = 100g$

  • $m_{\text{steam}} = 200g$

  • $T_{\text{water}} = 0°C$

  • $T_{\text{ice}} = 0°C$

  • $T_{\text{solid}} = 100°C$

  • $L_{1,2} = 334J/g$ (latent het from solid to liquid state)

  • $L_{2,3} = 2249J/g$ (latent heat from liquid to gas state)

1) Suppose $T_{final}$ = 0°C

Heat balance: $Q_{\text{Given By Steam}} = Q_{\text{Absorbed By Ice}}$

$m_{\text{solid}} L_{2,3} + m_{\text{solid}} C_{\text{water}}(T_{\text{solid}} - T_{\text{water}}) = m"_{\text{ice}}L_{1,2}$

after some calculations: $m"_{\text{ice}} = 1597,4g$

We conclude this supposition is wrong. We only have 200g of ice so there is no way 1597,4g which can melt.

2) Suppose $T_{\text{final}}=100°C$

Heat balance: $Q_{\text{Given By Steam}} = Q_{\text{Absorbed By Ice AND Water}}$

$m"_{\text{solid}}L_{2,3} = m_{\text{ice}}L_{1,2} + (m_{\text{ice}} + m_{water})c_{water}(T_{\text{solid}}-T_{\text{water}})$

after some calculations: $m"_{\text{solid}} = 126,5g$

This is not in contradiction with our supposition.


We can now tell that the final composition is 73,5g of steam, 726,5g of water and no ice.

My questions

  1. At the first supposition we suppose (and that is what is being used for the calculations):

$Q_{\text{given by steam}} = Q_{\text{absorbed by ice}}$

Why isn't it: $Q_{Given By Steam} = Q_{Absorbed By Ice AND Water}$ (like in the second supposition) as the ice and the water are both in thermic equilibrium?

  1. Why is at the first supposition the formula $Q=mc\Delta t$ associated with the heat given by the steam and at the second supposition with the heat absorbed by the ice and water?

Thank you very much,


1 Answer 1


It is simpler to first assume the water stays at 0 C while the ice melts (in equilibrium, that is what will happen). Once all the ice has melted, the remaining heat can be used to heat up the total mass of water - both that which started out as liquid, and that which started out as ice.

The second supposition asks whether there is enough heat to warm all the water to 100 C and still have heat left over in the steam. The answer is yes - again, this is just to make it easier to deal with the discontinuities that you would otherwise have in the equations. Perhaps a simple diagram makes this clear. See if you can calculate the breakpoints you would need here. Starting with a mixture of ice and water, we add heat (x axis). First, the ice disappears; eventually, some of the water evaporates. Does that help?

enter image description here

  • $\begingroup$ answer to question 1: Because if the water absorbs any energy, it will heat up above 0°C, which violates the premise. Only the ice can absorb energy and stay at 0°C. answer to Q2: still unclear. $\endgroup$ Commented Nov 1, 2016 at 8:44
  • $\begingroup$ Regarding point 2: the steam will initially cool to 100 C - there it can give off a lot of heat without changing temperature. In principle you could have hot steam at stage 2 but in fact you don't - the ice already took care of that. So the thing changing temperature is the water. $\endgroup$
    – Floris
    Commented Nov 1, 2016 at 12:24

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