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

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 5.830 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

total heat lost by ice = heat gained by steam

Now,

Let the final temperature be 'T'

amount of ice = 619 g

molar mass of water = 18 gm/mole

mass of steam present = (5.83*18) = 104.94 gm

Latent heat of melting of ice = 334 J/g

Specific heat capacity of, ice = 2.108 J/g-K

Specific heat capacity of water = 4.187 J/g-K

latent heat of vaporisation = 2260 J/g
Specific heat capacity of steam =: 1.996 J/g-K

Heat gained by water = heat required for temperature change from -12.4 to 0 + heat required for melting of ice + heat gained for rise in temperature from 0 to T

=> heat gained = (619*334)+(619*(2.108)*(0-(-12.4))) + 619*4.187*(T-0)
= (206746 + 1304.852*(12.4)) + 2591.753*T

= (222926.165 + 2591.753*T) Joules

Heat lost by steam = heat lost for change in temperature from 365 to 100 + heat lost during vaporisation + heat lost from change in temperature from 100 to T

=> heat lost = (104.94)*(1.996)*(365 - 100) + (104.94*2260) + 104.94*1.996*(100-T)

= (313617.387 - 209.46*T ) Joules

So,

(222926.165 + 2591.753*T) = (313617.387 - 209.46*T )

=> 2801.213*T = 90691.222

=> T = 32.375 degree C = final temperature

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