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mo You heat a 5.2 gram lead ball (specific heat capacity Joules - deg) to . You then drop the ball into 34.5 milliliters of water (density ; specific heat capacity ) at . What is the final temperature of the water when the lead and water reach thermal equilibrium?
temperature:

Ask by Rogers Beck. in the United States
Mar 24,2025

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Answer

The final temperature of the water is approximately .

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Bonus Knowledge

To solve for the final temperature of the water when thermal equilibrium is reached, we can apply the principle of conservation of energy. The heat lost by the lead ball will be equal to the heat gained by the water.
The formula for heat transfer is:
where is mass, is specific heat capacity, and is the change in temperature.
  1. For the lead ball:
    • Mass ( ) = 5.2 g
    • Specific heat capacity ( ) = 0.128 J/g-°C
    • Initial temperature ( ) = 183 °C
    • Final temperature ( ) = ? (same as water at equilibrium)
    Heat lost by lead:
  2. For the water:
    • Volume = 34.5 mL (which equals 34.5 g, since the density is 1 g/mL)
    • Specific heat capacity ( ) = 4.184 J/g-°C
    • Initial temperature ( ) = 22.4 °C
    Heat gained by water:
At thermal equilibrium:
Plug in the values:
Now, solving for :
Calculating the heat lost by the lead:
Calculating the heat gained by the water:
Now set them equal to each other:
Distributing:
Combining terms:
Now divide to find :
But since temperature cannot be this low, let’s re-check for potential error. The wrong step often stems from mathematics errors in the coefficients.
  • Final step result:
    Upon correcting, the real final temperature at equilibrium is close to-

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