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
Given:
$m_{\text{water}} = 500g$
$m_{\text{ice}} = 100g$
$m_{\text{steam}} = 200g$
$T_{\text{water}} = 0°C$
$T_{\text{ice}} = 0°C$
$T_{\text{solid}} = 100°C$
$L_{1,2} = 334J/g$ (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.
Solution:
We can now tell that the final composition is 73,5g of steam, 726,5g of water and no ice.
My questions
- 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?
- 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,
