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

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The discussion centers on calculating the minimum work needed to cool 5 kg of water from 20°C to 4°C using heat pumps. The key concept involves understanding the relationship between entropy change and the work required for this cooling process. Participants are seeking guidance on how to apply the change in entropy formula for a constant pressure scenario. The specific heat capacity of water is provided as 4184 J/kg°C, which is crucial for the calculations. The conversation emphasizes the importance of entropy in determining the efficiency of the cooling process.
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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 celcius, is cooled at constant pressure to 4 degree celcius 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?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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