Question

A hot cup of coffee is 200F when poured and placed in a room that is 70F. 10 min later the temperature of the coffee has cooled to 180F. What is the temperature of the coffee 10 min later?
(Assume Newton's Law of Cooling)

Answer 1

Answer:

The temperature of coffee becomes 163.08 Fahrenheit.

Step-by-step explanation:

The Newton's law of cooling can be mathematically written as

$-\frac{dT}{dt}=k(T-T_{room})\\\\\frac{-dT}{(T-T_{room})}=kdt$

Applying the limits in the above relation as

At t = 0 ; T= 200 F

At t =10 ; T = 180 F

$-\int_{200}^{180}\frac{dT}{T-70}=\int_{0}^{10}kdt\\\\-ln(T-70)_{200}^{180}=k\times 10\\\\\therefore k=\frac{ln(1.181)}{10}\\\\\therefore k=0.0167$

Applying the limits in the above relation as

At t = 10 ; T= 180 F

At t =20 ; T = T F

$-\int_{180}^{T}\frac{dT}{T-70}=\int_{10}^{20}0.0167dt\\\\-ln(T-70)_{180}^{T}=0.167\\\\\therefore T = 163.08F$

Answer 2

Answer:

The temperature of coffee becomes 163.08 Fahrenheit.

Step-by-step explanation: