Ice at -14.1°C and steam at 117°C are brought together in an insulated container until thermal equilibrium is reached at 51.6°C. Using the heat capacities and latent heats of ice and steam, as well as the initial and final temperatures, the ratio of the mass of steam to the mass of ice is calculated to be 0.232.
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Ice at -14-1 C and steam at 117 C are brought together at atmospheric.docx
1. Ice at -14.1 °C and steam at 117 °C are brought together at atmospheric pressure in a
perfectly insulated container. After thermal equilibrium is reached, the liquid phase at 51.6 °C
is present. Ignoring the container and the equilibrium vapor pressure of the liquid at 51.6 °C,
find the ratio of the mass of steam to the mass of ice. The specific heat capacity of steam is 2020
J/(kg C°).
Solution
Heat lost by steam =Heat gained by ice
Q steam =Q ice
m s c s (117 - 100) +m s L v +m s c w (100-51.6) =m I C i (0-(-14.1)) +m i L f +m i c w (51.6-0)
m s (2020*17 +2.26*10 6
+4186*48.4] =m i [2000*14.1 +3.35*10 5
+4186*51.6]
m s /m i = 0.232