Copyright

A cup of tea, originally at 95°C, cools down according to Newton's Law of Cooling:...

Question:

A cup of tea, originally at 95°C, cools down according to Newton's Law of Cooling:

{eq}\frac{dH}{dt} = \alpha (H - A) {/eq}

where H(t) is the tea temperature at time t, α is a negative constant, and A = 20°C is the temperature of the surrounding air. Suppose that after 10 minutes, the tea temperature is 65°C.

1. Find explicitly H(t).

2. Calculate the maximal length, i.e. {eq}lim_{t \rightarrow \infty} H(t) {/eq}.

Newton's cooling Law

The rate of change in the temperature of the object is proportional to the difference between its temperature and temperature of the environment. It is given by the differential equation {eq}\displaystyle \frac{dH}{dt} = \alpha (H - A) {/eq} which can be solved using separation of variables. The actual function that describes the temperature as a function of time can be solved by applying the initial conditions.

Answer and Explanation:


The change in temperature of the cup of tea is described by the differential equation

{eq}\displaystyle \frac{dH}{dt} = \alpha (H - A) {/eq}

wher...

See full answer below.

Become a Study.com member to unlock this answer! Create your account

View this answer

Learn more about this topic:

Loading...
What are Heating and Cooling Curves?

from College Chemistry: Help and Review

Chapter 6 / Lesson 5
69K

Related to this Question

Explore our homework questions and answers library