Question

The half-life of caffeine in a healthy adult is 4.8 hours. Jeremiah drinks 18 ounces of caffeinated coffee in the morning. How long will it take for only 60% of the caffeine to remain in his body? O In 5 3 In 2 t = 5 3 In In 2 t= (4.8) (18) t= In In 2 302 (4.8) In 3 5 In 2 352 (18)

Answer 1

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Answer 2

Correct Question

The half-life of caffeine in a healthy adult is 4.8 hours. Jeremiah drinks 18 ounces of caffeinated coffee in the morning. How long will it take for only 60% of the caffeine to remain in his body?

(a). $t=(\frac{ln\frac{5}{3} }{ln\frac{1}{2} } ) (4.8)$

(b). $t=(\frac{ln\frac{3}{5} }{ln\frac{1}{2} } ) (4.8)$

(c). $t=(\frac{ln\frac{5}{3} }{ln\frac{1}{2} } ) (18)$

(d). $t=(\frac{ln\frac{3}{5} }{ln\frac{1}{2} } ) (18)$

Answer:

It will  take  $t=(\frac{ln\frac{3}{5} }{ln\frac{1}{2} } ) (4.8)$ hours for only 60% of the caffeine to remain in his body and option (b) is correct.

Step-by-step explanation:

Given:

The half-life of caffeine in a healthy adult is 4.8 hours.

Jeremiah drinks 18 ounces of caffeinated coffee in the morning.

60% of the caffeine to remain in his body.

To Find:

How long will it take for only 60% of the caffeine to remain in his body.

Formula Used:

The half-life means the time required for a quantity to reduce  to half of its initial value.It is represented by symbol $t_{\frac{1}{2} }$.

$N(t)=N(i)(0.5)^{\frac{t}{t_{\frac{1}{2} } } }$   ----------- formula no.01.

$N(t)$ = the remaining amount of substance after time t.

$N(i)$ = the initial amount of substance.

Solution:

As given,the half-life of caffeine in a healthy adult is 4.8 hours.

The half-life of caffeine in a healthy adult, $t_{\frac{1}{2} } =4.8 \ hours$.

As given,Jeremiah drinks 18 ounces of caffeinated coffee in the morning.

$N(i) = 18 \ ounces$.

As given,60% of the caffeine to remain in his body.

The remaining amount of substance after time t, $N(t)$ =60% of the 18.

                                                                                        $=18\times\frac{60}{100} \\=10.80 \ Ounces.$

Applying the formula no.01.

$N(t)=N(i)(0.5)^{\frac{t}{t_{\frac{1}{2} } } }$      

$10.80= 18(0.5)^{\frac{t}{4.8 } } }\\\frac{10.80}{18} = (0.5)^{\frac{t}{4.8 } } }\\ \frac{3}{5} =(\frac{1}{2} )^{\frac{t}{4.8 } } }$

Taking log of both sides.

$ln\frac{3}{5} =\frac{t}{4.8}\ ln\frac{1}{2}$

$t=(\frac{ln\frac{3}{5} }{ln\frac{1}{2} } ) (4.8)$ hours.

Thus, It will  take  $t=(\frac{ln\frac{3}{5} }{ln\frac{1}{2} } ) (4.8)$ hours for only 60% of the caffeine to remain in his body and option (b) is correct.

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