A 0.5530-kg ice cube at-12.40°C is placed inside a chamber of steam at 365.0. La

Question

A 0.5530-kg ice cube at-12.40°C is placed inside a chamber of steam at 365.0. Later, you notice that the ice cube has completely melted into a puddle of water. If the chamber initially contained 6.790 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.) Number

Explanation / Answer

here ,

let the final temperature is T

specific heat

for ice 2.108 kJ/kgK, and for water vapor (steam) 1.996 kJ/kgK

heat lost by steam = heat gain by ice

0.5530 * (12.40 * 2108 + 334 *10^3 + 4186 * T ) = 6.790 * 0.018 * ((365 -100) * 1.996 + 2260 * 10^3 + 4186 * (100 - T))

solving for T

T = 45.4 degree C

the final temperature is 45.4 degree C

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