What is the final equilibrium temperature when 20.00 grams of ice at 15.0°C is mixed with 5.000 grams of steam at 120.0°C?
The specific heat of ice is 2.100 kJ/kg °C, the heat of fusion for ice at 0°C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, the heat of vaporization of water at 100.0°C is 2,256 kJ/kg, and the specific heat of steam is 2.020 kJ/kg °C.
The specific heat of ice is 2.100 kJ/kg °C, the heat of fusion for ice at 0°C is 333.7 kJ/kg, the specific heat of water 4.186 kJ/kg °C, the heat of vaporization of water at 100.0°C is 2,256 kJ/kg, and the specific heat of steam is 2.020 kJ/kg °C.
1 Answer
Answer:
Explanation:
The question doesn't state the phase of the mixture at equilibrium. Due to the presence of the latent heat of phase change, it is necessary to start with an assumption of the final state of the mixture and verify the accuracy of the assumption after determining the temperature of the final mixture.
Assuming that the mixture is in its liquid state at its equilibrium temperature
A wellinsulated system shall see no heat exchange with the surroundings. Thus energy conserves within the system.
for which

#"E"("released") = "E"("condensation") + "E"("cooling")#
#color(white)("E"("released")) = "L"_color(purple)(v) * m("steam")#
#color(white)("E"("released") =) + c("steam") * m("steam") * (120 °C  100 °C)#
#color(white)("E"("released") =) + c("water") * m("steam") * (100 °C color(navy)( t) °C)# 
#"E"("absorbed") = bb() bb("(")"E"("fusion") + "E"("heating")bb(")")#
#color(white)("E"("absorbed")) = bb() bb("(")"L"_color(purple)(f) * m("ice")#
#color(white)("E"("absorbed") =) + c("ice") * m("ice") * ((15) °C  0 °C)#
#color(white)("E"("absorbed") =) + c("water") * m("ice") * (0 °C color(navy)( t) °C)bb(")")#
Equating
The value of