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A 235 g lead ball at a temperature of 83.4 ∘C is placed in a

light calorimeter containing 156 g of water at 21.8 ∘C.
Find the equilibrium temperature of the system.

1 Answer

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Answer:

To find the equilibrium temperature of the system, we can apply the principle of energy conservation, assuming no heat is lost to the surroundings. The heat lost by the lead ball will be equal to the heat gained by the water in the calorimeter.

The heat gained or lost can be calculated using the formula:

Q = mcΔT

Where Q is the heat gained or lost, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

For the lead ball:

Q₁ = m₁c₁ΔT₁

For the water:

Q₂ = m₂c₂ΔT₂

Since the total heat lost by the lead ball is equal to the total heat gained by the water, we can set Q₁ = Q₂ and solve for the equilibrium temperature.

m₁c₁ΔT₁ = m₂c₂ΔT₂

Given:

m₁ = 235 g (mass of the lead ball)

c₁ = specific heat capacity of lead (0.13 J/g⋅°C)

ΔT₁ = equilibrium temperature - initial temperature of the lead ball

m₂ = 153 g (mass of the water)

c₂ = specific heat capacity of water (4.18 J/g⋅°C)

ΔT₂ = equilibrium temperature - initial temperature of the water

Substituting the values into the equation:

235 g * 0.13 J/g⋅°C * (equilibrium temperature - 81.9 °C) = 153 g * 4.18 J/g⋅°C * (equilibrium temperature - 22.3 °C)

Simplifying and solving for the equilibrium temperature:

30.55 (equilibrium temperature - 81.9) = 638.94 (equilibrium temperature - 22.3)

30.55 equilibrium temperature - 30.55 * 81.9 = 638.94 equilibrium temperature - 638.94 * 22.3

30.55 equilibrium temperature - 2522.345 = 638.94 equilibrium temperature - 14227.342

-608.39 equilibrium temperature = -11749.997

equilibrium temperature ≈ 19.31 °C

Therefore, the equilibrium temperature of the system is approximately 19.31 °C.

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