Answer:
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.
