2.0 kg of solid gold (au) at an initial temperature of 727 °c is allowed to exchange energy with 1.5 kg of liquid gold at an initial temperature of 1063 °c. the solid and liquid can only exchange energy with each other. when the two reach thermal equilibrium, will the mixture be entirely solid, or will there be a mixed solid/liquid phase? explain how you know

The mixture will be entirely solid.

Explanation:

Mass of solid gold: ms=2.0kg   Solid Initial tremperature: Tos= 727°C    (Absorbs heat)

Mass of liquid gold: ml=1.5kg   Liquid Initial tremperature: Tol= 1063°C (Release heat)

Two bodies reach thermal equilibrium when the heat transmitted is equal to the heat absorbed and both reach the same temperature (equilibrium temperature):

Q= Heat

CeAu= Specific heat of gold

Q absorbed = Q released

Te= equilibrium temperature

ms*CeAu*(Te-Tos)= ml*CeAu*(Tol-Te)

You can simplify the specific heat of gold, and we have:

ms*(Te-Tos)= ml*(Tol-Te)

Applying distributive property:

ms*Te - ms*Tos = ml*Tol - ml*Te

We place the terms "Te" on the right side of equality and on the left side of equality the known data:

ms*Te + ml*Te = ml*Tol + ms*Tos

"Te" as a common factor:

Te*(ms +ml) = ml*Tol + ms*Tos

Te = (ml*Tol + ms*Tos)/ (ms +ml)

Te= (1.5kg*1063°C  + 2.kg *727°C)/(2.0kg+1.5kg)

Te= 3,048.5kg*°C/3.5kg

Te= 871°C

melting point of gold = 1,064 °C

At 871°C gold is solid

and 1and 1