Answer:
In this case, the final temperature of both the lead weight and the water at thermal equilibrium is approximately 52.17°C.
Step-by-step explanation:
To find the final temperature of both the lead weight and the water at thermal equilibrium, we can use the principle of heat transfer. Heat flows from a hotter object to a colder object until they reach the same temperature.
Let's assume the final temperature is T (in degrees Celsius). The heat gained by the lead weight is equal to the heat lost by the water. We can use the equation:
m₁* c₁ * (T - T₁) = m₂ * c₂ * (T₂ - T)
where:
- m₁ is the mass of the lead weight (2.00 g),
- c₁ is the specific heat capacity of lead (0.128 J/g°C),
- T₁ is the initial temperature of the lead weight (11.0°C),
- m₂ is the mass of the water (7.92 g),
- c₂ is the specific heat capacity of water (4.18 J/g°C), and
- T₂ is the initial temperature of the water (52.6°C).
Substituting the values into the equation, we get:
(2.00 g) * (0.128 J/g°C) * (T - 11.0°C) = (7.92 g) * (4.18 J/g°C) * (52.6°C - T)
Simplifying the equation, we have:
0.256(T - 11.0) = 33.0576(52.6 - T)
Now, let's solve for T:
- 0.256T - 2.816 = 1735.4632 - 33.0576T
- 0.256T + 33.0576T = 1735.4632 + 2.816
- 33.3136T = 1738.2792
- T ≈ 52.17°C
Therefore, the final temperature of both the lead weight and the water at thermal equilibrium is approximately 52.17°C.