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

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Homework Help Overview

The problem involves a thermal interaction between a sample of lead shot and water, specifically calculating the final temperature of the mixture when lead at a high temperature is introduced to water at a lower temperature. The subject area encompasses concepts from thermodynamics and heat transfer.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the principle of heat transfer, recalling that the heat lost by the lead should equal the heat gained by the water. They express confusion about how to relate heat capacity to their calculations. Other participants mention the need to consider phase changes if the final temperature exceeds 100°C.

Discussion Status

The discussion includes attempts to clarify the relationship between heat transfer and specific heat capacities. Some participants have provided reminders about relevant equations and considerations regarding phase changes, indicating a productive exploration of the topic.

Contextual Notes

Participants note the specific heats of lead and water, as well as the mass of each substance involved. There is an acknowledgment of the potential phase change of water at 100°C, which adds complexity to the problem.

zhen
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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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