How many joules of heat are needed to raise the temperature of 30.0 g of aluminum from 22°C to 80°C, if the specific heat of aluminum is 0.90 J/g°C?

There are 1566 joules of heat needed to raise the temperature of 30.0 g of aluminum from 22°C to 80°C, if the specific heat of aluminum is 0.90 J/g°C.

Explanation:

Given: Mass = 30.0 g

Specific heat =

Formula used to calculate the heat energy requires is as follows.

where,

q = heat energy

m = mass of substance

C = specific heat capacity of substance

= initial temperature

= final temperature

Substitute the values into above formula as follows.

Thus, we can conclude that there are 1566 joules of heat needed to raise the temperature of 30.0 g of aluminum from 22°C to 80°C, if the specific heat of aluminum is 0.90 J/g°C.

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