A 0.6190-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.

Question

A 0.6190-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 6.070 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

Answer – We are given , mass of ice = 0.619 kg = 619 g , ti = -12.40oC

Moles of steam = 6.070 moles , ti = 365oC , tf = ?

Mass of steam = 6.070 moles * 18.015 g/mol

                        = 109.35 g

So the heat loss form the steam is gain by the ice

q ice = m *C ice * t

q steam = - m *C steam * t

so, q ice = q steam

so, m *C ice * t = - m *C steam * t

619 g * 2.018 J/goC * (tf –(-12.40))oC = - 109.35 * 1.996 J/goC * (tf –350)oC

So, 1249.14 tf + 218.3 tf = 76392.6 - 15489.4

1467.04 tf = 60903.28

So, tf = 41.50oC

final temperature of the puddle once it settled to equilibrium is 41.50oC

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