How Is Entropy Related to the Minimum Work Needed to Cool Water?

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SUMMARY

The discussion centers on calculating the minimum work required to cool 5 kg of water from 20°C to 4°C using heat pumps, with a specific heat capacity (Cp) of 4184 J/kg·°C. The key concept involves understanding the change in entropy during this cooling process. The expression for the constant pressure entropy change is crucial for determining the minimum work, which can be derived from the formula ΔS = m * Cp * ln(T2/T1) for the temperature change.

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metalboyxp
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This question buzz my head :cry: Plz help
A mass of water (mass m=5kg, specific heat capacity at constant pressure: Cp=4184j/kg*C) initially in thermal equilibrium with the atmosphere at 20 degree celsius, is cooled at constant pressure to 4 degree celsius by means of heat pumps operating between water & atmosphere. What is the minimum work required? :confused:
The hint for this question is that using the change in entropy, but how?
 
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what is the expression for constant pressure entropy change?
 

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