A calorimeter is filled with 0.76 kg of water $(c = 4186 \, \text{J/kg°C})$ at $17°C$. If a 0.368 kg piece of lead $(c = 128 \, \text{J/kg°C})$ is heated to $91°C$ and placed in the water, what is the equilibrium temperature?

The heat gained by the water is equal to the heat lost by the heated lead in a calorimeter. We can use this relationship and the specific heat and mass of water and lead to find the equilibrium temperature.

Explanation:

The subject of this question is about calorimetry and equilibrium in a system. The heat gained by the water in the calorimeter will be equal to the heat loss by the lead. Hence, we can use the formula Q = mcΔT, where m is mass, c is specific heat, and ΔT is the change in temperature.

For the lead, Q(lead) = m c ΔT => Q(lead) = 0.368 kg * 128 J/kgC * (91oC - Teq) where Teq is the equilibrium temperature.

And for water Q(water) = m c ΔT => Q(water) = .76 kg * 4186 J/kgC * (Teq - 17oC).

Since the heat gained by water is equal to the heat lost by lead, we can derive the equation, Q(lead) = Q(water), and solve it to find the equilibrium temperature.

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A calorimeter is filled with 0.76 kg of water \((c = 4186 \, \text{J/kg°C})\) at \(17°C\). If a 0.368 kg piece of lead \((c = 128 \, \text{J/kg°C})\) is heated to \(91°C\) and placed in the water, what is the equilibrium temperature?

Answer :

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