PHS A 0.5970-kg ice cube at -12.40degree C is placed inside a chamber of steam a
ID: 2271533 • Letter: P
Question
PHS
A 0.5970-kg ice cube at -12.40degree C is placed inside a chamber of steam at 365.0degree C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 6.190 moles of steam (water) molecules before the ice is added, calculate the final temperature of the puddle once it settled to equilibrium. (Assume the chamber walls are sufficiently flexible to allow the system to remain isobaric and consider thermal losses/gains from the chamber walls as negligible.)Explanation / Answer
we know, molar mass of water, M = 18 gram/mole
no of moles, n = mass/molar mass
= m/M
==> mass = n*M
mas of steam, m1 = 6.19*18
= 111.42 grams
= 0.11142 kg
mass of ice, m2 = 0.553 kg
Let T is the final temperature.
Apply,
Heat last by steam = heat gained by ice
m1*C_steam*(365 - 100) + m1*Lv + m1*C_water*(100-T) = m2*C_ice*(0 - (-12.4)) + m2*Lf + m2*C_water*(T - 0)
0.11142*2010*265 + 0.11142*2.245*10^6 + 0.11142*4186*100 - 0.11142*4186*T = 0.553*2030*12.4 + 0.553*3.33*10^5 + 0.553*4186*T
356126 - 466.4*T = 198069 + 2314.86*T
T*(466.4 + 2314.86) = 356126 - 198069
T = (356126 - 198069)/(466.4 + 2314.86)
= 56.8 degrees celsius
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