Does doubling boiling water change the time to reach thermal equilibrium with ice?

[mengshuen]

Sorry to trouble all of you here again, but

Imagine that there is 0.50kg of 0 degree ice added to 1kg of boiling water.

after a few minutes they achieve thermal equilibrium.

however, instead of adding 1kg of boiling water, now I add 2kg of boiling water, keeping the mass of the ice constant, will they take a different time to reach thermal equilibrium, taking into account that the temperature after thermal equilibrium is achieved is different?

 

[Astronuc]

One needs to show some work or effort.

A key point - keeping the mass of the ice constant - means constant heat transfer area. Heat flux is proportional to temperature difference - that it still the same.

What is different is the equilibrium temperature and the total energy in the system.

Is the thermal time constant variant or invariant with respect to the total energy or total mass in the system?

 

[mengshuen]

I am guessing that it will take a different time to reach thermal equilibrium, hence the thermal time constant would be variant to total energy in the system.

Are there any equations for me to find out the rate of thermal energy transfer/loss to the ice, with only the mass, and temperature difference known? The question also requires one to calculate what would be the difference in time to achieve thermal equilibrium.

 

[Astronuc]

Well one way would be to 'look at it from the perspective of the ice'.

The mass of ice is the same! In one case the mass must achieve one temperature, while in the second case, it must get to a higher temperature.

The key then is to show that the rate of heat flow into the ice is the same or not. If the heat flow is the same, then it takes longer to get to the higher temperature.

 

[mengshuen]

Can I say that the heat flow is the same, as in the relationship in thermal conductivity? If I understand correctly, the power given off by the boiling water is proportional to the temperature gradient. So now there is the same temperature gradient, and hence rate of energy transferred is the same. More energy is required to get the ice to a higher temp, so more time is needed.