A 0.6410-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.

Question

A 0.6410-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 5.950 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

mass of steam, m1 = 5.95*18

= 107.1 grams

= 0.1071 kg

mass of ice, m2 = 0.641 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)

0.1071*2010*265 + 0.1071*2.245*10^6 + 0.1071*4186*100 - 0.1071*4186*T = 0.641*2030*12.4 + 0.641*3.33*10^5 + 0.641*4186*T

342318 - 448.32*T = 229588 + 2683.2*T

T*(448.32 + 2683.2) = 342318 - 229588

T = (342318 - 229588)/(448.32 + 2683.2)

= 36 degrees celsius

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.