0.05625 m^3 of H2O(s) is added to a thermally isolated system of 20 L of H2O (l). If the ice was initially at 0 C and the liquid water was initially at 95 C, what is the final state and temperature?
3. The attempt at a solution
I missed class for this so I’m going by what I remember as my book doesn’t exactly cover this from what I can tell.
I looked up the density for solid water at 0 C and converted the volume to mass, there are 51.4 kg H2O (s)
I looked up the density for liquid water at 95 C and converted the volume to mass, there are 19.2 kg H2O (l)
Ice is going to gain energy as it goes from a phase change of solid to liquid. The liquid mass that was once ice is then going to gain energy as it reaches thermal equilibrium with the liquid water originally at 95 C. The source of the heat is going to be the liquid water. Since heat in = heat out,
q (phase change) + q (temp. raise) = q (temp. lower)
I calculated q (phase change) by looking up the heat of fusion of H2O which is 334 J/g.
q (phase change) = (334 J/g)(51,400g)
The specific heat of water is (4.184 J/g),
q (temp. raise) = (4.184 J/g)(51,400g) Δt
q (temp. lower) = (4.184 J/g)(19,200g) Δt
Putting it all together:
(334 J/g)(51,400g) + (4.184 J/g)(51,400g) Δt = (4.184 J/g)(19,200g) Δt
But this gives a Δt of 127.4 K which doesn’t make sense considering the initial temperatures. It’s been a while since I’ve covered this topic and can’t really tell what I’m doing wrong. Would anyone be able to point me in the right direction?