You order a 16 oz glass of tea (where the mass of water is 474 grams) from a local restaurant. the tea is freshly brewed and has an initial temperature of 28.29 °c. you add ice to cool it. if the heat of fusion of ice is 6.020 kj/mol and each ice cube contains exactly 1 mol of water, how many ice cubes are necessary to cool the tea to 0.41 °c? the specific heat of the "tea" is 4.184 j/g*c.

Explanation:

It is known that relation between heat energy, specific heat and temperature change is as follows.

                  Q =

where,    q = heat energy

              m = mass

              C = specific heat

          = change in temperature = =

Therefore, calculate the heat energy as follows.

               Q =

                   =

                  = 55292.06 J

or,              = 55.3 kJ        (as 1 kJ = 1000 J)

One ice cube is equal to 1 mole.

Hence, 6.020 kJ/mol of energy absorbed by 1 ice cube. And, number of ice cubes absorbing 55.3 kJ of energy will be calculated as follows.

                 = 9

Thus, we can conclude that 9  ice cubes are necessary to cool the tea to .

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

b. examine and describe cells