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A 2.00 g lead weight, initially at 11.0 ∘C , is submerged in 7.92 g of water at 52.6 ∘C in an insulated container.

What is the final temperature of both the weight and the water at thermal equilibrium?

User Ammad Khan
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6 votes

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.

User Yanill
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