A glass containing a mass of water m=.30kg at 60 degrees Celsius has a mass of i

Question

A glass containing a mass of water m=.30kg at 60 degrees Celsius has a mass of ice m=.025kg at its melting point of 0 degrees Celsius added to it. The mixture is allowed to come to equilibrium at a final temperature Tf. Constants : c(water) = 4186 J/kg*K, c(ice) = 2050 J/kg*K at 100 degrees Celcius, 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 * 333kJ/kg + 0.025 * 4.186kJ/kgK * T = 0.3kg * 4.186kJ/kgK * (60-T)

B)

0.025 * 4.186kJ/kgK * T + 0.3kg*4.186kJ/kgK * T = 0.3kg * 4.186kJ/kgK * 60 - 0.025kg * 333kJ/kg

1.36T = 67.023

T = 49.2653

Tf = 49.3 Celsius

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