Physical application: Low of cooling

Physical application: Low of cooling
The rate of change of temperature of a body is proportional to the difference between the temperature of a body and the temperature of the surrounding medium T_(s ). Suppose T is the temperature of the body at time t, then :
dT/dt=-k(T-T_s )
Example 1: Placed a metal bar, at a temperature of 100°F in a room with constant temperature of 0°F. After 20 minutes the temperature of the bar is 50°F. Determine:
1. the time required to reach the bar at a temperature of 25°F .
2.the temperature of the bar after 10 minutes.
Solution: T_s=0°F
dT/dt=-k(T-T_s )
dT/dt=-k(T-0 ) ? dT/dt=-kT (Separable varibles)
dT/T=-kdt
ln?T=-kt+c_1
T=e^(-kt+c_1 )=e^(-kt).e^(c_1 )
T=ce^(-kt) ; e^(c_1 )=c (General solution)
(Initial condition) t=0 ,T=100°F
100=ce^(-k×0) ? c=100
T=100e^(-kt)
To find k we have t=20 min , T=50°F?50=100e^(-k×20)
e^(-20k)=0.5 ? -20k=ln?(0.5)
? k=ln?(0.5)/(-20)=0.0347
T=100e^(-0.0347t)
1. t=? if T=25°F
25=100e^(-0.0347t)
e^(-0.0347t)=0.25 ? 0.0347t=ln?(0.25) ?t=40 min
2. t=10 min , T=?
T=100e^(-0.0347×10)=70.68?
Example 2: A body at an unknown temperature is placed in a room which is held at a constant temperature of 30° F. If after 10 minutes the temperature of the body is 0° F and after 20 minutes the temperature of the body is 15° F, find the unknown initial temperature.
Solution: T_s=30°F
dT/dt=-k(T-T_s )
dT/((T-30 ) )=-kdt
ln?(T-30)=-kt+c_1
T-30=e^(-kt+c_1 )
T=30+ce^(-kt) ;c=e^(c_1 )
t=10 ,T=0 ?0=30+ce^(-10k)
ce^(-10k)=-30 ?(1)
t=20 ,T=15 ?15=30+ce^(-20k)
ce^(-20k)=-15 ?(2)
(1)÷(2) ? e^10k =2
10 k=ln?2=0.693
k=0.0693
To find c ? ce^(-10×0.0693)=-30
c=-30×e^0.693=-60
T=30-60e^(-0.0693t)
t=0 ?T=30-60e^(-0.0693×0) ?T=-30° F







Electric Circuits
We will stady an electric circuits that consists of a voltage source, a resistor and an inductor. Consider an electric circuits RL, with an applied voltage E volts and constant resistance of R ohms and constant inductance of L henries. The current i=i(t) in amperes, satisfies the differential equation
L di/dt+Ri=E(t)
Example 3: In RL electric circuits . Find current i=i(t) at any time t.

Solution: L di/dt+Ri=E
0.25 di/dt+15i=12 ? di/dt+60i=48
di/(48-60i)=dt ? ?_0^i?di/(48-60i)=?_0^t?dt
-? ? ( 1 )/60 ln?|48-60i| ?|_0^i=t?|_0^t
-( 1 )/60 (ln?|48-60i|-ln??48 ? )=t ? ln?|48-60i|-(ln?48 )=-60t
ln??(48-60i)/48?=-60t ? (48-60i)/48=e^(-60t) ? 1-( 5 )/( 4 ) i=e^(-60t)
i=( 4 )/( 5 ) (1-e^(-60t) )



( 1 ) A body at a temperature of 50° F is placed in an oven whose temperature is kept at 150° F. If after 10 minutes the temperature of the body is 75° F, find the time required for the body to reach a temperature of 100° F. (ans : 24.094 minutes)
(2) In RL electric circuits, if E=10v ,L=0.5H ,R=12?. Find current
i=i(t) at any time t.