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A 1 kg brick of lead at 300°C is placed in a 10 kg tank of water at 20°C. The system is thermally insulated. Find the equilibrium temperature of the lead and the water.

Specific heat of lead = 128 J/kg·K

Answer :

The equilibrium temperature of the lead and water system is approximately 94.6°C.

Initially, the heat gained by the lead equals the heat lost by the water until they reach thermal equilibrium. We can use the principle of conservation of energy, which states that the heat lost by the lead (Q_lead) is equal to the heat gained by the water (Q_water).

To calculate Q_lead, we use the formula:

Q_lead = mcΔT

Where m is the mass of lead, c is the specific heat capacity of lead, and ΔT is the change in temperature. Substituting the given values, we get:

Q_lead = (1 kg)(128 J/kg•K)(T - 300°C)

Similarly, for the water, using the same principle:

Q_water = mcΔT

Q_water = (10 kg)(4186 J/kg•K)(T - 20°C)

Equating Q_lead and Q_water, we get:

1^128*(T - 300) = 10^4186*(T - 20)

Solving this equation for T, the equilibrium temperature, we find T ≈ 94.6°C.

This result indicates that the system will reach equilibrium at approximately 94.6°C, where the lead and water are in thermal equilibrium, having exchanged heat until their temperatures equalize.

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