Question

The temperature of a cup of coffee is 85oC initially, and becomes 75°C after 1 min. Estimate the temperature after 5 min. Assume the room temperature is 25°C. Is it still warm enough to drink?

Answer 1

Answer:

$T(5)=49.11^\circ C$

Well, In my opinion, it's still warm enough to drink

Explanation:

Let's use Newton's Law of Cooling  given by:

$T(t)=T_a+(T_o-T_a)*e^{-kt}$    (1)

Where:

$T(t)=Temperature\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}coffee\hspace{2 mm}at\hspace{2 mm}time\hspace{2 mm}t\hspace{2 mm}(in\hspace{2 mm}minutes)$

$T_o=Initial\hspace{2 mm}temperature\hspace{2 mm}of\hspace{2 mm}the\hspace{2 mm}coffee=85^\circ C$

$Ta=Ambient\hspace{2 mm}temperature=25^\circ C$

$k=Time\hspace{2 mm}constant$

$t=Time\hspace{2 mm}in\hspace{2 mm}minutes$

Replacing the data provide in (1)

$T(t)=25+(85-25)*e^{-kt}= 25+60*e^{-kt}$ (2)

So, we need to find k. But we know:

$T(1)=75$

Using that information in (2)

$T(1)=25+60*e^{-k} =75$ (3)

Solving for k in (3)

Sustract 25 to both sides:

$60*e^{-k} =50$

Multiply both sides by $\frac{1}{60}$

$e^{-k} =\frac{5}{6}$  

Express $e^{-k}$ as $\frac{1}{e^{k}}$

$\frac{1}{e^{k}}=\frac{5}{6}$

Natural logarithm to both sides:

$ln(\frac{1}{e^{k} })=ln(\frac{5}{6} )\\ ln(1)-ln(e^{k})=-0.1823215568\\0-k=-0.1823215568\\k=0.1823215568$

Replacing k in (2)

$T(t)= 25+60*e^{-(0.1823215568)t}$ (4)

Evaluating t=5 in (4)

$T(5)= 25+60*e^{-(0.1823215568)5}=25+60*e^{-0.911607784} =49.11265432^\circ C$

I attached you the graph of the function (4)