Answer to Question #264542 in Differential Equations for mary

Question

1. A radioactive substance plutonium 239 has a half life of 24100 years. Initially, there is 30mg of the substance, find how much will remain for the first 1000 years. How long for the substance to decay 90% of its initial mass? Ans. 29.15mg; 80,058.47 years
2.  A thermometer reading of 50°C was plunged into a tub of frozen water. If the thermometer reads 30° after 5 seconds, what will be the reading after 10 seconds? How long will it take for the thermometer reading to drop to 20°C? Ans. 18°C; 8.97 seconds

Answer 1

1.

$R=R_0e^{kt}$

A half life is 24100 years

$\dfrac{1}{2}R_0=R_0e^{k(24100)}$

$k(24100)=-\ln 2$

$k=-\ln 2/24100$

$R=R_0(2)^{-t/24100}$

Initially $R_0=30 mg$.

(i)

$R(1000)=300(2)^{-1000/24100}$

$R(1000)=29.15mg$

(ii)

$R=R_0(2)^{-t/24100}=0.1R_0$

$\ln((2)^{-t/24100})=\ln(0.1)$

$-t/24100=\ln (0.1)/\ln(2)$

$t=24100(\ln(10)/\ln(2))$

$t=80058.467\ years$

$t=80058.47\ years$

2. The temperature of a body at any time $t$ is

$T_b=(T_{b0}-T_m)e^{kt}+T_m$

Initially, ܶ$T_{b0}=50\degree C$ and $T_m=0\degree C$

Given $T_b(5)=30\degree C$

$30=(50-0)e^{5k}+0$

$e^{5k}=0.6$

(i)

$T_b(10)=(50-0)e^{10k}+0=50(0.6)^2$

$T_b(10)=18\degree C$

(ii)

$k=0.2\ln(0.6)$

$T_b(t_1)=(50-0)e^{0.2\ln(0.6)t_1}+0=20$

$e^{0.2\ln(0.6)t_1}=0.4$

$t_1=\dfrac{5\ln(0.4)}{\ln(0.6)}$

$t_1=8.97\ sec$