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 6.310 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

Here ,

specific heat of steam , Ss = 1996 J/(kg * degree C)

mass of steam , ms = 6.310 * 0.018 Kg

ms = 0.114 Kg

latent heat of fusion = 334 *10^3 J/Kg

latent heat of vapourization = 2260 *10^3 J/Kg

let the final temperature is T

heat lost by steam = heat gain by ice

.6410 * (2108 * (12.4) + 4186 * (T) + 334 * 10^3) = 0.114 *(1996 * (365 - 100) + 2260 * 10^3 + 4186 *(100 - T) )

solving for T

T = 42.7 degree C

the final temperature of water puddle is 42.7 degree

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