A 0.6630-kg ice cube at -12.40°C is placed inside a chamber of steam at 365.0°C.
ID: 1484740 • Letter: A
Question
A 0.6630-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 6.430 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
heat required by ice to get converted to water = mL + msdT
= 0.6630 *1000 * 334 + 0.6630*1000 * 2.03 * 12.40
= 238131.036 J
mass of water = 6.430*18 =115.74 gm
latent heat of vaporisation = 2264.76
heat contained in water = (115.74)*1.996*(365-100) + (115.74)* ( 2264.76 ) + ((115.74))*4.187 *100
=371803.176 J
remaining energy after converting all ice to water
371803.176 -238131.036
=133672.14 J
let final temprature be T
(115.74 +0.6630*1000) 4.187 dT = 133672.14
T = 40.9963750118 deg
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