A camper pours 500 g of coffee, initially in a pot at 80 oC, into a 110-g aluminum cup initially at 30 oC. What is the final temperature? Assume that coffee has the same specific heat as water and that no heat is exchanged with the surroundings. (cwater=4.18 J/goC). (cAluminum=0.900J/g. C)

The final temperature : t=77.74 °C

Given

500 g coffee, t = 80 °C

110 g aluminum cup, t = 30 °C

Required

The final temperature

Solution

Q in = Q out

Q absorbed = Q released

Q aluminum cup = Q coffee

110 x 0.9 x (t-30)=500 x 4.18 x (80-t)

99t-2970=167200-2090t

2189t=170170

t=77.74 °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

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