The element Unobtainium has a half-life of 3 years. Let M(t) be the mass of Unobtainium at time t starting with an initial amount of 14 kg. (a) Give a formula for M(t). (b) After how many years will the initial mass of Unobtainium shrink to 1 kg? Round your answer to one digit after the decimal point. 6. Starbucks serves coffee at 170° and the room temperature in Starbucks is 70° The coffee cools to 100° after 10 minutes. Let T(t) be the temperature of the coffee at time 1, measured in minutes, (a) Write down the differential equation for T(t) and determine a formula for T(t). (b) From the time when the temperature is 100° at t = 10, how many additional minutes will it take for the temperature of the coffee to reach 80°? Round your answer to che digit after the decimal point.

Part 1

a) M(t) = 14 e⁻⁰•²³¹ᵗ

b) 11.4 years

Part 2

a) The differential equation for T is

(dT/dt) = -k(T - T∞)

(dT/dt) = -k(T - 70)

And the solution of the differential equation

T(t) = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

b) The additional minutes it will take for the temperature of the coffee to reach 80°C after reaching 100°C = 9.1 minutes.

Step-by-step explanation:

Part 1

Radioactive reactions always follow a first order reaction dynamic

Let the initial concentration of Unobtainium be M₀ and the concentration at any time be M

(dM/dt) = -kM (Minus sign because it's a rate of reduction)

(dM/dt) = -kM

(dM/M) = -kdt

 ∫ (dM/M) = -k ∫ dt 

Solving the two sides as definite integrals by integrating the left hand side from M₀ to M and the Right hand side from 0 to t.

We obtain

In (M/M₀) = -kt

(M/M₀) = e⁻ᵏᵗ

M(t) = M₀ e⁻ᵏᵗ

Although, we can obtain k from the information on half life.

For a first order reaction, the rate constant (k) and the half life (T(1/2)) are related thus

T(1/2) = (In2)/k

T(1/2) = 3 years

k = (In 2)/3 = 0.231 /year.

And M₀ = 14 kg from the question, M(t) becomes

M(t) = 14 e⁻⁰•²³¹ᵗ

b) What is t, when M(t) = 1 kg

M(t) = 14 e⁻⁰•²³¹ᵗ

1 = 14 e⁻⁰•²³¹ᵗ

e⁻⁰•²³¹ᵗ = (1/14) = 0.07143

-0.231t = In (0.07143) = - 2.639

t = 2.639/0.231 = 11.4 years

Part 2

a) Let T be the temperature of the coffee at any time

T∞ be the temperature of the room = 73°C

T₀ be the initial temperature of the coffee = 170°C

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the coffee = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

T(t) = T∞ + (T₀ - T∞)e⁻ᵏᵗ

Inputting the known variables

T(t) = 70 + (170 - 70) e⁻ᵏᵗ

T(t) = 70 + 100e⁻ᵏᵗ

But it is given that, at t = 10, T = 100°C

100 = 70 + 100e⁻¹⁰ᵏ

100e⁻¹⁰ᵏ = 30

e⁻¹⁰ᵏ = 0.3

-10k = In (0.3) = 1.204

K = 0.1204

T(t) = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

b) T(t) = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

We first calculate the time it takes to get 80°C

80 = 70 + 100e⁻⁰•¹²⁰⁴ᵗ

100e⁻⁰•¹²⁰⁴ᵗ = 10

e⁻⁰•¹²⁰⁴ᵗ = 0.1

-0.1204t = In (0.1) = -2.303

t = (2.303/0.1204) = 19.1 minutes.

From the time it reached 100°C, i.e. t = 10 minutes, 19.1 minutes is 9.1 minutes extra.

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