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.

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years

answer; natural resources;