Question

Warm air is less dense than cold air. The same is true of water. You decide to take a bath and turn on both the hot and cold faucets. When the tub is one-third full, you dip your hand in and discover the water is too cold. You turn the cold water faucet off and keep the hot water turned on. You leave the hot water running for ten minutes. When you return and dip your hand in, you discover that the water in the tub is different temperatures in different places. It is too hot close to the faucet and in a top layer across the entire tub. When you sit down, however, the water in the bottom of the tub is cold. Explain how this happens and why: • how the hot water ended up in a layer on the top and near the faucet. • why this is or is not a transfer of heat. • how the temperature of the water in the tub will change over time. • why you will eventually want to get out of the tub, considering the air and your porcelain enameled steel tub.

Answer 1

Given that the warm water is less dense than the cold water, we have;

• The cold water which is more dense than the warm water will remain settled in the lower part of the tub while particles of the less dense warm water will move above the cold water not being able to penetrate lower due of difference in their gravitational potential energies

• This is a case of (lack of) heat transfer due to inability of heat to penetrate into the lower layers and transfer the heat by conduction, from particle to particle, and therefore, one layer, the top remains warm while the other layer, the lower layer, remains cold
• With time (over an hour) more and more warm water units will be able to penetrate deeper and make the lower layer warmer while the warm water at the top loses heat by evaporation to the atmosphere, until the entire water reaches equilibrium
• Heat will continue to be lost by evaporation to the air and by conduction through the porcelain to the steel tub until the water in the tub becomes cold

Step-by-step explanation: