Asked by KS
Find the mass of water that vaporizes when 4.04 kg of mercury at 210 °C is added to 0.498 kg of water at 81.6 °C.
I am stuck on trying to figure out how to solve this. Please help!
I am stuck on trying to figure out how to solve this. Please help!
Answers
Answered by
Anonymous
vaporization takes place while the Hg is above 100C, or 100C+.
so until the water heats to 100, no vaporization.
HeatGained by water+heatgainedHg=0
.498*cwater*(100-81.6)+4.04*CHg*(Tf-210)=0
so calculate Tfinal for the Hg...note it is not in equilibrium with the water temp...
Now, with the water at 100C, and Hg still hot at Tf.
sume of heats gained=0
heatinsteam+heatgainedbyHg=0
Mass*HvWater+.498*Chg*(100-Tfabove)=0
solve for mass...
so until the water heats to 100, no vaporization.
HeatGained by water+heatgainedHg=0
.498*cwater*(100-81.6)+4.04*CHg*(Tf-210)=0
so calculate Tfinal for the Hg...note it is not in equilibrium with the water temp...
Now, with the water at 100C, and Hg still hot at Tf.
sume of heats gained=0
heatinsteam+heatgainedbyHg=0
Mass*HvWater+.498*Chg*(100-Tfabove)=0
solve for mass...
Answered by
Anonymous
I'm still not getting the right answer for it...
Answered by
Steve
so, what are your calculations? Maybe we can figure out what's wrong.