Differential Equation - a few. (1 Viewer)

[skillz]

Hey guys, can someone give me a hand.
The rate of decay of a radioactive substance is proportional to the amount Q of matter present at any time, t. The differential equation for this situation is dQ/dt = -kQ where k is a constant. If Q = 50 when t=0 and Q=25 when t= 10 , find the time taken for q to reach 10.

What i did was..
dq/dt = -kq
dt/dq= -1/kq
= -1/k x (1/q)
integrate, t= -1/k loge Q + C
when t=0 q =50
0= -1/k (loge50) + c
loge50 = c

when t=10 q=25
k10 = -loge (25) + loge 50
k=~ -.069
therefore.
5= -1/-.069 x (1/ 10)

but it does not equal the answer ?

2nd q.
If the thermostat in an electric heater failed, the rate of increase of temperature , dtheta/dt, would be .01 theta degrees per minute where theta is in degrees Kelvin (K) and t is in minutes. If a heater was switched on at a room temperature of 300 K and the thermostat did not function, what would the temperature of the heater be after 10 minutes?

nfi :|


Cheers BOS

 

[pLuvia]

1.
dQ/dt=-kQ
t=-lnQ/k+C
When t=0 Q=50 C=ln(50)/k
t=-lnQ/k+ln(50)/k
When t=10 Q=25
10=1/k[-ln(25)+ln(50)]
k=[ln2]/10

When Q=10
t=-ln(10)/k+ln(50)/k
=23.219..
=23.22(2dp)

 

[Riviet]

2. I think this is how you do it:
Let's let theta=T for simplicity.
Given dt/dT=0.01T=T/100,
dT/dt=100/T
.'. t=100lnT + C by integrating both sides with respect to t
when t=0 and T=300,
C=-100ln300
.'. t=100lnT - 100ln300
t=100ln(T/300)
when t=10
ln(T/300)=10/100
T/300=e^1/10
T=300.e^1/10
~332

.'. temperature of heater after 10 minutes is 332K.