Sir Isaac Newton found that the temperature of something heated will cool down according to the function:

and

- Use the info about how long it takes for my coffee to get to find
*k*

So the constant of cooling for my coffee is -.1627 or so. Here is the check:*u(t) = Room Temp + (Heated Temp-Room Temp)e*^{kt}

150 = 72 + (180-72)*e*^{k2}

78 = (108)*e*^{2k}

*e*= 78/108 = 13/18^{2k}

2*k*=ln(13/18)

*k*=.5(ln(13/18)) ≈ .5(-.3254224)=-0.1627112*u(t)*= 72+(108)*e*^{-.1627t}*u*(2)=72+108*e*^{(-.1627)(2)}≈150° - Solve for
*t*#### 72+108

*e*^{-.1627t}= 120°

108*e*^{-.1627t}= 48

*e*^{-.1627t}= 48/108=4/9

-1627*t*= ln(4/9)

*t*≈ (-.8109302)/(-.1627) = 5 minutes

- In a 72° room, my 180° coffee will be 150° after two minutes. How long will it take to get 75°?
- What is the temperture after 30 minutes?
- Boiling water (212° at sea level) is left in a 70° and after 5 minutes it is
180° What is the constant of cooling?
- Using this info from the previous question, how long will it take to have it cool to 98°?
- Heating is cooling in reverse. Use the same constant
*k*as in #3. If an ice cube is placed in the same room. how long will it take to become 50°? (presume the ice is 32° when frozen). - What temperature is the water after 15 minutes?

Here is a link to my java code

Here is a link to other examples of exp and log functions from the UBC