A quantity m=.025kg of steam at its boiling point of 100 degrees Celcius is mixe
ID: 2168894 • Letter: A
Question
A quantity m=.025kg of steam at its boiling point of 100 degrees Celcius is mixed with m=.4kg of water at an initial temperature of 50 degrees Celsius and allowed to come to equilibrium at a final temperature Tf. Constants : c(water) = 4186 J/kg*K, c(steam) = 4219 J/kg*K, Latent heat of fusion = 333 kJ/kg, Latent heat of vap = 2256 kJ/kg.
a) write the Q equation for the equilibrium condition. (Write is as a single equation ) Use symbols for everything except the temperatures.
Hint : Q_gain = Q_loss
b) Solve this equation for the final temperature.
Explanation / Answer
A)
0.025kg*2256 kj/kg + 0.025kg*4.186 kJ/kgK *(50-T) = 0.4kg *4.186 kJ/kgK * T
B)
0.4kg *4.186 kJ/kgK * T + 0.025kg*4.186 kJ/kgK * T = 0.025kg * 2256 kj/kg + 0.025kg*4.186 kJ/kgK *50
1.77905T = 61.6 K
T = 34.64 K
Tf = 50 + 34.64 = 84.6 Celsius