A quantity m=.025kg of steam at its boiling point of 100 degrees Celcius is mixe

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

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