Question

What is the half-life of the goo in minutes? Find a formula for G(t), the amount of goo remaining at time t. How many grams of goo will remain after 68 minutes?

Answer 1

To solve this question on the half-life, we will use this expression:

`\begin{gathered} G(t)=G_oe^(-kt) \\ \text{where G(t) is the remaining sample at time t.} \\ G_{o\text{ }}\text{ is the original sample} \\ K\text{ is a constant} \\ t\text{ is time} \end{gathered}`

To proceed in solving, we will need to find the value of constant k

`\begin{gathered} G(t)=G_oe^(-kt) \\ G(t)=17.25 \\ G_o=276 \\ t=255 \\ \text{Now substitute the parameters above into the formula:} \\ 17.25=276e^(-k(255)) \\ (17.25)/(276)=e^(-k(255)) \end{gathered}`
`\begin{gathered} 0.0625=e^(-k255) \\ \ln 0.0625=-255k \\ (\ln 0.0625)/(-255)=k \\ 0.0109=k \end{gathered}`

Now to get the half-life in minutes will be to get the time taken for the sample to go from 276g to 138g.

`\begin{gathered} G(t)=G_oe^(-kt) \\ G(t)=138g \\ 138=276e^(-0.0109t) \\ (138)/(276)=e^(-0.0109t) \\ 0.5=e^(-0.0109t) \\ \ln 0.5=-0.0109t \\ (\ln 0.5)/(-0.0109)=t \\ 63.591\text{minutes = t} \end{gathered}`

The half-life is 63.59 minutes.

The formula for G(t) at time t is:

`G(t)=276e^(-0.0109t)`

The amount of goo that will remain after 68 minutes is calculated using the formula above:

`\begin{gathered} G(t)=276e^(-0.0109t) \\ t=68\text{ minutes} \\ G(t)=276e^(-0.0109(68)) \\ G(t)=276e^(-0.7412) \\ G(t)=131.5255\text{ grams} \\ G(t)\text{ = 131.53 grams (to 2 d.p)} \end{gathered}`

The amount of goo remaining after 68 minutes is 131.53 grams.