Math homework question

Question

A pot of boiling soup with an internal temperature of 100° Fahrenheit was taken off the stove to cool in a 65°F room. After fifteen minutes, the internal temperature of the soup was 96°F.

To the nearest minute, how long will it take the soup to cool to 82°F?

Answer 1

The formula for Newton's Law of Cooling is:

�(�)=��+(�0−��)⋅�−��T(t)=Ta​+(T0​−Ta​)⋅e−kt

Where:

1. �(�)T(t) is the temperature of the object at time �t.
2. �0T0​ is the initial temperature of the object.
3. ��Ta​ is the ambient temperature (temperature of the surroundings).
4. �k is a constant that depends on the properties of the object and its surroundings.
5. �e is the base of the natural logarithm.

Given the information:

�0=100°�T0​=100°F (initial temperature of the soup) ��=65°�Ta​=65°F (temperature of the room) After 15 minutes, �(15)=96°�T(15)=96°F.

Substituting these values into the formula:

96=65+(100−65)⋅�−15�96=65+(100−65)⋅e−15k

Solving for �k, we get:

�−15�=3135e−15k=3531​

�=−115⋅ln⁡(3135)k=−151​⋅ln(3531​)

Now, we can use this value of �k to find out how long it will take for the soup to cool to 82°F:

82=65+(100−65)⋅�−��82=65+(100−65)⋅e−kt

�−115⋅ln⁡(3135)�=1735e−151​⋅ln(3531​)t=3517​

Taking the natural logarithm of both sides:

−115⋅ln⁡(3135)�=ln⁡(1735)−151​⋅ln(3531​)t=ln(3517​)

�=15⋅ln⁡(1735)ln⁡(3135)≈26.8t=15⋅ln(3531​)ln(3517​)​≈26.8

Rounded to the nearest minute, it will take the soup about 27 minutes to cool to 82°F.