If 2 cups of ice at room temperature began to melt, would the ice in the cup where the resulting water was drained (or drunk) melt faster than the ice in the cup that retained all the resulting water?
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Anonymous
ANSWERS: 2
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The one that resulting water drained. When the surrounding of ice is water after some point the equilibrium will be fixed at zero degree for a while but this system is not totally closed we have air surface of this system and this air in a way also effect the temperature of this system and will lose eq. and will melt totally and at the final it will reach the temperature of the room. But when there is no water the ice will always be forced to reach room temperature. Because there is two component ice and air and the mass of air is way bigger than ice mass. "Thermodynamic 0th rule" wil rule
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I constructed a great argument then lost it. I'll try again. Energy emission rates depend on (Tobject - T surroundings)^4, it's a black body spectra emission thing, essentially. The thing to remember is that the ice is in a cup, and the air will chill with the ice's increase in temperature. Temperature transfers are via kinetic energy changes, e.g. collisions. Air is significantly less dense than water, and the temperature difference will make very little affect on the actual energy of the particles... I think what will happen is that the water will melt it faster, as it provides a full surface interaction versus the air which doesnt... Not to mention the air will also cool and will form a 'protective layer' around the ice, much like putting your hand in liquid nitrogen and why it doesn't freeze instantly. Find out and let me know :)
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