Can Freezing and Melting Water Warm a Home? Answers Here!

  • Thread starter Thread starter Danko
  • Start date Start date
  • Tags Tags
    Temperature
AI Thread Summary
Freezing and melting water can slightly warm a home, but the effect is minimal due to air's poor heat conductivity. When a freezer operates, it exhausts more heat into the room than the energy used to freeze the water, primarily due to inefficiencies. However, the total warmth generated cannot exceed the electrical energy consumed, as the heat extracted from the water returns when it melts. If the ice melts outside, the heat from the freezer remains in the house, potentially raising the temperature slightly. Overall, while there may be a small increase in warmth, it will not surpass the energy input from the freezer.
Danko
Messages
2
Reaction score
0
Hi I'm having some problems with a question i have to present in class.

If you want to raise the temperature at home, will this work?:
You put a bucket of water in the freezer, remove it when it has frozen and let it melt outside.

Will the house get warmer, and is it possible that more warmth has been created than the electrical energy that has been used.

Can anyone help?
 
Physics news on Phys.org
Danko said:
Hi I'm having some problems with a question i have to present in class.

If you want to raise the temperature at home, will this work?:
You put a bucket of water in the freezer, remove it when it has frozen and let it melt outside.

Will the house get warmer, and is it possible that more warmth has been created than the electrical energy that has been used.

Can anyone help?

Yes,the room will definitely warm,but not too much...Air is very bad heat conductor and the quantity of heat will be very small.

For the second part,no...

Daniel.
 
What are the assumptions here? If you turn on the freezer just to freeze the water, then yes, the freezer will exhaust more heat than is required to melt the ice afterwards. The freezer, like any machine, is not perfectly efficient: it uses slightly more energy than is required. The heat taken out of the water is exhausted into the room along with the additional heat from friction, etc. When you take the ice out of the freezer and let it melt, some of that heat will go into the water but no more than came from the freezing. The heat from the inefficiency of the freezer will remain.

However, you can NOT get more energy out that is put in: you cannot get more warmth than corresponds to the electrical energy. (The heat that was initially in the water and was exhausted from the freezer is back in the water after it melts.)
 
thanks

Thanks for the help!
I think the idea was (although I'm not sure) that the ice was to melt outside the house, in that case shouldn't the warmth that the freezer emits remain in the house thus resulting in a slightly higher temperature?

Secondly about the electrical power...Does the freezer consume any more power than what is needed to power the condenser? Anything proportional to the amount of water?

Also i think you are to assume that the freezer is on already.
 
Danko said:
Will the house get warmer, and is it possible that more warmth has been created than the electrical energy that has been used.
I find myself disagreeing with the other two posters on this second point.

I agree that the house will heat up by freezing the ice. But it will warm up by more than the electrical energy used.

The amount that it will warm up will be the heat extracted from the ice plus the work done by the refrigerator (which ultimately is dissipated as heat). So the total heat will necessarily be greater than the amount of work done (ie. power consumed) by the refrigerator (ie. by the amount of heat extracted).

AM

PS. Taking it outside would not necessarily cause it to melt. Not where I am today (-28 deg. C)
 
Last edited:
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}...
Back
Top