What is the Final Temperature When Lead Shot is Mixed with Water?

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SUMMARY

The final temperature when a 1000 g sample of lead shot at 300°C is mixed with 100 g of water at 5.6°C can be calculated using the principle of conservation of energy. The specific heats are 0.129 J/g°C for lead and 4.184 J/g°C for water. The energy calculations yield 38700 J for lead and 2343.04 J for water. If the final temperature exceeds 100°C, the heat of vaporization of water, 2.26 J/kg, must be considered due to the phase change.

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The question is:
A 1000 g sample of lead shot, at 300°C, is dropped into 100 g of water at a temperature of 5.6°C. The specific heats of lead and water are 0.129 and 4.184 J/g°C, respectively. What is the final temperature of the mixture, in degrees Celsius?

I was doing my chemistry pratice, then I saw this question, I don't have a clue at all. It looks very similar to the high school physics, but I can not remember anything. Can you just give me some hints?

I just remember the change of Tw and Tl is the same.
but how does it relate to the heat capacity?...

I solve for the energy of the water: 100 X 4.184 X 5.6 = 2343.04 J
.......lead: 1000 X 0.129 X 300 = 38700 J

then what should I do?:frown:

please help,
 
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nevermind, i finially remember what to do...
 
And if the final temperature exceeds 100°C, there is a phase change of the water at 100°C, which would also have to be considered.

Heat of vaporization of water = 2.26 J/kg or 539 cal/g.

Hopefully you remembered \Delta H= m cp\Delta T
 
Astronuc said:
And if the final temperature exceeds 100°C, there is a phase change of the water at 100°C, which would also have to be considered.
Heat of vaporization of water = 2.26 J/kg or 539 cal/g.
Hopefully you remembered \Delta H= m cp\Delta T

I remembered it after I post, but I really appriciated it, thank you
 
Last edited:

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