How Much Dry Steam Is Required to Melt Ice in an Open System?

AI Thread Summary
To determine the amount of dry steam required to melt 450 kg of ice at -18 degrees Celsius and achieve a final effluent temperature of 5 degrees Celsius, one must consider the energy transfer involved in the process. The scenario is classified as an open system due to the addition of steam mass. The necessary calculations involve determining the heat required to raise the ice temperature to 5 degrees and equating it to the heat lost by the steam as it cools. Although specific enthalpy values are not provided, the problem can be approached using algebraic expressions for heat transfer. The solution involves setting up and solving the equations for heat gained by the ice and heat lost by the steam.
chopper
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I need a kick start for the following question. We're studying open/closed systems at the moment.

"450kg of ice at -18 degrees C is to be removed from a freezer using a jet of steam. How much dry steam at 100 degrees C is required if the final temp of the effluent is 5 degrees C?"

This is what I've sussed out so far.. because mass is being added to the system (the mass of the steam) I assume this is an open system problem. I haven't been given any information regarding enthalpy of either the ice or the steam so I'm assuming it's a simple matter of working out how much energy I need to bring the ice up to 5 degrees and introducing that energy into the freezer in the form of steam that will be giving up energy as it cools down to 5 degrees.

Can anyone point me in the right direction?

Cheers
Chopper
 
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chopper said:
I need a kick start for the following question. We're studying open/closed systems at the moment.

"450kg of ice at -18 degrees C is to be removed from a freezer using a jet of steam. How much dry steam at 100 degrees C is required if the final temp of the effluent is 5 degrees C?"

This is what I've sussed out so far.. because mass is being added to the system (the mass of the steam) I assume this is an open system problem. I haven't been given any information regarding enthalpy of either the ice or the steam so I'm assuming it's a simple matter of working out how much energy I need to bring the ice up to 5 degrees and introducing that energy into the freezer in the form of steam that will be giving up energy as it cools down to 5 degrees.
It should be a simple matter of algebra. Just set up the expression for heat h needed to raise the temperature of 450 kg of ice to 5 degrees and the heat lost by a mass m of stream. Equate the two and solve for m.

AM
 
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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