A 0.5750-kg. ice cube at -12.40 degree C is placed inside a chamber of steam at
ID: 1573306 • Letter: A
Question
A 0.5750-kg. ice cube at -12.40 degree C is placed inside a chamber of steam at 365.0 degree 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
given
m_Ice = 0.575 kg
m_steam = n*M
= 5.83*18*10^-3
= 0.105 kg
we know,
C_Ice = 2100 J/(kg K)
L_Ice = 3.33*10^5 J/kg
C_water = 4186 J/(kg K)
L_steam = 2.26*10^6 J/kg
C_steam = 1.996*10^3 K(kg K)
let T is the final equilibrium temperature.
use heat hained by ice = heat lost by water
m_Ice*C_Ice*(0-(-12.4)) + m_Ice*L_ICe + m_Ice*C_water*(T - 0 ) = m_steam*C_steam*(365-100) + m_steam*L_steam + m_steam*C_water*(100-T)
0.575*2100*12.4 + 0.575*3.33*10^5 + 0.575*4186*T = 0.105*1996*265 + 0.105*2.26*10^6 + 0.105*4186*(100-T)
==> T = 45.8 degrees celsius