A 2.43 g lead weight, initially at 10.4 degrees Celsius, is submerged in 7.74 g

Question

A 2.43 g lead weight, initially at 10.4 degrees Celsius, is submerged in 7.74 g of water at 52.3 degrees Celsius in an insulated container.
What is the final temperature of both the weight and the water at thermal equilibrium? Express the temperature in Celsius to 3 significant figures. A 2.43 g lead weight, initially at 10.4 degrees Celsius, is submerged in 7.74 g of water at 52.3 degrees Celsius in an insulated container.
What is the final temperature of both the weight and the water at thermal equilibrium? Express the temperature in Celsius to 3 significant figures.
What is the final temperature of both the weight and the water at thermal equilibrium? Express the temperature in Celsius to 3 significant figures.

Explanation / Answer

Call the final temperature Tf. Then the change in temperature

specific heat capacity of water 4186 J/(kg Co).

Q1 = [4186 J/(kg Co)] [7.74 kg] [Tf - 52.3oC]

Q2 = [4186 J/(kg Co)] [2.43kg] [Tf - 10.4oC]

Q1 + Q2 = 0

[4186 J/(kg Co)] [7.74 kg] [Tf - 52.3oC] + [4186 J/(kg Co)] [2.43kg] [Tf - 10.4oC] = 0

Both of these materials are essentially water so their two specific heats are the same. Since the specific heats are the same, we can immediately factor that out and reduce this equation to

[7.74 kg] [Tf - 52.3oC] + [2.43kg] [Tf - 10.4oC] = 0

7.74Tf - 404.8 + 2.43Tf - 25.27 = 0

10.17 Tf = 430.07

         Tf = 42.2880C

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