Thermodynamics Problem: Calculating Mass of Lead and Water in Equilibrium

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The discussion revolves around a thermodynamics problem involving the equilibrium of molten lead and water. A worker pours 1.25 kg of molten lead at 327.3°C into 0.500 kg of water at 75°C, and the goal is to calculate the mass of lead and water remaining after reaching thermal equilibrium. The calculations involve specific heat capacities and phase transitions, with the final temperature determined to be approximately 355.57°C. It is noted that some water will boil away, affecting the final mass, and the importance of considering phase changes for both lead and water is emphasized. The discussion highlights the need for clarity in defining the remaining mass of water versus steam in the final answer.
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Homework Statement



Not sure if this is the right thread for this question but here goes :

A worker pours 1.25 kg of molten lead at a temperature of 327.3 into 0.500 kg of water at a temperature of 75 C in an insulated bucket of negligible mass. Assuming no heat loss of the surroundings. Calculate the mass of lead and water remaining in the bucket when the materials have reached thermal equilibrium.

Homework Equations





The Attempt at a Solution



C(water)=4190
C(lead ) = 130

Q=mcdelta T

(1.250)(130)(600.3-t)= Q (lead)

(1.250)(4.19)(T-348)=Q(water)

expand rearrange :

97500-162.5T=5237.5T-1822650

T= 355.57333

i kind of lost here , that's all i have done

the answear is

(1.743kg) - remember, steam will be produced!

but how is this the answear can someone show me the working ?

thanks
 
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What happens to water when it reaches a temperature of 100 C? Hint: think teapot.
 
SteamKing said:
What happens to water when it reaches a temperature of 100 C? Hint: think teapot.

boiling point of water ? .

but how is the mass achieved ?
 
You have used the mass of lead (1.25 kg) instead of the mass of water (0.500 kg) in writing the equation for the amount of heat received by the water.
 
Start out by assuming that not all the water boils away, but that there is enough heat in the lead to raise the temperature of the water to its boiling point. Under these circumstances, what is the final temperature of the lead and water?
 
Take into consideration that the melting point of lead is 600 K (327.3oC) so the lead must go a phase transition to solid
 
Lead changing to water or vice-versa? Alchemy perhaps?
If all that's left is lead and water then steam must count as water ...

Or maybe lead oxidizes or ... ?
 
The problem states "molten lead at 327.3 oC"

My textbook has the melting point of lead to be 600.3 K. A phase transition (liquid to solid) for the lead is necessary, as well as phase transition from liquid to vapor for some of the water.

From where are you coming with an alchemy reference?

The quoted "answer" is obtained when the phase transitions are taken into consideration.
 
PHd Austin said:
The problem states "molten lead at 327.3 oC"

My textbook has the melting point of lead to be 600.3 K. A phase transition (liquid to solid) for the lead is necessary, as well as phase transition from liquid to vapor for some of the water.

From where are you coming with an alchemy reference?

The quoted "answer" is obtained when the phase transitions are taken into consideration.

The mass of lead obviously doesn't change, so 50% of the answer is a no-brainer if there ever was one. It starts and ends at 1.25 kg. Easy, huh?

If we decide not to call steam "water" then obviously one can ask for the proportion of water to steam. The problem should have been stated as "what is the remaining proportion of water to vapor".

Alchemy is the changing of one element to another, which would have to happen if the mass of lead were to chage - unless, as I said, we consider chemical reactions like oxidation.
 
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