A 7.5 gram sample of zinc (specific heat = 0.390 J/g°C) is heated and then placed in a calorimeter containing 65.0 grams of water. The temperature of water increases from 20.00°C to 25.00°C. What was the initial temperature of the zinc?

489.88 was the initial temperature of the zinc.

Explanation:

Given :

mass of zinc = 7.5 grams

mass of water = 65 grams

specific heat = 0.390J/g°C

specific heat of water = 4.184 J/g°C

initial temperature of water = 20 degrees

Final temperature of zinc and water will be same as zinc is dissolved in water thus T2zinc and T2 water = 25 degrees

ΔTzinc = change in temperature T2 - T1

initial temperature of zinc =?

The formula used is

q = mcΔT for both water and zinc

putting the values for zinc and water in the formula:

mzinc x czinc x T2 zinc - T1 zinc = mwater x c water x T2 water- T 1 water

7.5 x 0.390 x ΔT zinc = -(65 x 4.184 x (25-20)

2.925 x  ΔT zinc  = -1359.8

ΔT zinc  = - 464.88 degrees

Now,

ΔT zinc = T2 zinc - T1 zinc

T1 zinc = - 464.88-20

=489.88 degrees

489.88 degrees is the initial temperature of the zinc.

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

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa