Differential equation help pls?

Question

The rate at which the temperature X of a liquid increases is inversely proportional to its temperature at time t. Write a differential equation and solve it.

Given that the temperature of the liquid increases from 20^C to 20.5^C in 2seconds.

find

1. temperature of body after 10seconds

2. the further time for the temperature to increase frm 20.5^C to 21^C.

Thanks

Answer 1

From the above we Know:

1.     dT/dt   =   K/t         (where T is the temperature

                                                    t  is time

                                                    K is the constant of

                                                        proportionality)

Hence the differential equation is:

         dT   =   K dt / t

    => delta T    =   K ln( t2 / t1)  ........ (1)  [ where 'delta'

                                                                     stands for change]

But we know that ,

          During the temperature rise from 20^C to 20.5^C

         From (1)

         0.5   =  K ln { (t+2) / t }

          Thus K  =  1 / 2ln { (t+2) / t } ...........(2)

Now to Calculate Change in temperature after 10 seconds

      delta T  =  [ln { (t+10) / t }] / 2 [ ln { (t+2) / t } ]

Thus the temperature after (t + 10) seconds is

Tt   +   delta T     ANSWER 1

Now for Change in time from 20.5^C to 21^C

Here using equation (1) we can write

0.5  =  ln { ( t+2   +   delta t ) / (t + 2) } /  2 ln ( t + 2 / t)  

OR     1  =   ln { ( t+2   +   delta t ) / (t + 2) } / ln ( t + 2 / t)  

=>     ln ( t + 2 / t)  =   ln { ( t+2   +   delta t ) / (t + 2) }

=>  t + 2 / t   =   ( t + 2  +  delta t ) / ( t + 2)

=>  delta t    =   2 ( 1 + 2/t)    ANSWER 2

Answer 2

dT/dt = k/T

T*dT = k*dt

T^2/2 = kt + C

Set t= 0 to be when T = 20C

So we have two points: t = 0 and T = 20 / t = 2 and T = 20.5

(20^2)/2 = k(0) + C

C = 200

(20.5^2)/2 = k(2) + 200

k = 5.0625

After 10 seconds:

T^2/2 = (5.0625)(10) + 200

T = 22.39C after 10 seconds

Time to reach 21C.

(21^2)/2 = (5.0625)t + 200

20.5 = (5.0625)t

t = 4.05 seconds

Additional time to reach 21C is 4.05 - 2 or 2.05 seconds

Answer 3

dT/dt = k/T

T*dT = k*dt

T^2/2 = kt + C

Set t= 0 to be when T = 20C

So we have two points: t = 0 and T = 20 / t = 2 and T = 20.5

(20^2)/2 = k(0) + C

C = 200

(20.5^2)/2 = k(2) + 200

k = 5.0625

After 10s:

T^2/2 = (5.0625)(10) + 200

T = 22.39C after 10s

Time to reach 21C.

(21^2)/2 = (5.0625)t + 200

20.5 = (5.0625)t

t = 4.05s

Additional time to reach 21C is 4.05 - 2 or 2.05s