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A 3.6-kg piece of aluminum at 28.8 ∘C is placed in 1.0 kg of water in a Styrofoam container at room temperature (20.0 ∘C).

Estimate the net change in entropy of the system.

User Iain Brown
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The net change in entropy for the system, comprising aluminum and water, is 1.2256 J/°C.

To calculate the net change in entropy of the system, we need to find the entropy change of the aluminum and the water separately, and then add them together. The formula to calculate the entropy change is ΔS = mCΔT, where ΔS is the entropy change, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

For the aluminum, we can use the equation ΔS = mCΔT. The mass of the aluminum is 3.6 kg, the specific heat capacity of aluminum is 0.897 J/g°C, and the change in temperature is (24 - 28.8) °C. Plugging these values into the equation, we get ΔS = (3.6 kg)(0.897 J/g°C)(-4.8 °C).

For the water, we can use the same equation ΔS = mCΔT. The mass of the water is 1.0 kg, the specific heat capacity of water is 4.18 J/g°C, and the change in temperature is (24 - 20) °C. Plugging these values into the equation, we get ΔS = (1.0 kg)(4.18 J/g°C)(4.0 °C).

Finally, we can add the entropy changes of the aluminum and water together to get the net change in entropy of the system. ΔS_net = ΔS_aluminum + ΔS_water = (3.6 kg)(0.897 J/g°C)(-4.8 °C) + (1.0 kg)(4.18 J/g°C)(4.0 °C).

For aluminum:

ΔS_aluminum = (3.6 kg) × (0.897 J/g°C) × (-4.8°C)

ΔS_aluminum = -15.4944 J/°C

For water:

ΔS_water = (1.0 kg) × (4.18 J/g°C) × (4.0°C)

ΔS_water = 16.72 J/°C

Net change in entropy:

ΔS_net = ΔS_aluminum + ΔS_water

ΔS_net = -15.4944 J/°C + 16.72 J/°C

ΔS_net = 1.2256 J/°C

So, the net change in entropy for the system is 1.2256 J/°C.

User Fendi Tri Cahyono
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