Question

200g water at 0 degree centigrade mixed with 200g ice at -25 degree centigrade temperature.find the ratio of mass water and ice after mixing.

Answer 1

Let's suppose that we have ice and water mixture after reaching heat ballance, which is possible only at temperature of 0 celsium. If we have another situation, we will have impossible numbers in equation and will change assumption. Heat equation in this case:

$λm_w=c_im_i\cdot25$, where $m_w$ is mass of frozen water, and $m_i$ is mass of ice. If the water is 0 degree and is freezing, the ice can`t melt. When it reaches 0 degree we will have a system ice-water at 0 degree, which is stable. Thus, $333500\cdot m_w=2050\cdot0.2 \cdot 25 \Rightarrow m_w=\frac{2050\cdot 0.2\cdot 25}{333500}\approx0.031kg$.

So, 32 gramms of water will freeze and ratio of water and ice masses is $\frac{200-31}{200+31}=0.73$.