Three identical beakers each hold 1000 g of water at 20 ∘c. 100 g of liquid water at 0∘c is added to the first beaker, 100 g of ice at 0 ∘c is added to the second beaker, and the third beaker gets 100 g of aluminum at 0∘c. the contents of which container end up at the lowest final temperature?

The second beaker

Explanation:

This is a technical question, so let's answer it with some technical definitions.

First, let's take a look at the specific heat of the water, ice and aluminum:

Aluminum: 0.9

Water: 4.18

Ice: 2.10

Now, we can see that aluminum has the smallest value of all. because of this, once the aluminum is dropped into the beaker the change of temperature will be also smallest as it's not releasing too much heat. So this will have the highest temperature of the three beakers.

Now in the case of beaker 1 with water at 0°C and beaker 2 with ice at 0°C,  the ice by logic, has to melt first into the water. In order to do this, it use latent heat from the water in the beaker, so it takes more energy from that. In beaker 1 these event do not occur because we have water at the same state (both liquid) so, they only mix and the temperature is low, but in beaker 2, onoy because ice has to melt first, and take latent heat from water, this lowers even more the temperature. Therefore, the beaker with ice has the lowest final temperature.