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

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

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

Step-by-step 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
`(1)/(60)`

`e^(-k) =(5)/(6)`

Express
`e^(-k)` as
`(1)/(e^(k))`

`(1)/(e^(k))=(5)/(6)`

Natural logarithm to both sides:

`ln((1)/(e^(k) ))=ln((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)