Question

When a cold drink is taken from a refrigerator, its temperature is 5°C. After 25 minutes in a 20°C room its temperature has increased to 10°C. (Round your answers to two decimal places.) (a) What is the temperature of the drink after 45 minutes? °C (b) When will its temperature be 16°C? min

Answer 1

Answer:

a)T=12.77°C

b) t=83.38 min

Explanation:

$T_i$=5°C

After 25 min, T=10°C  

Air temperature $T_s$=20°C

We know that cold drink will cool because its surrounding temperature is high .

From Newton's law of cooling

$\dfrac{dT}{dt}= K(T-T_s)$

Where K is the constant,T is the temperature at any time t and $T_s$ is the surrounding temperature.

$\dfrac{dT}{dt}=K(T-20)$

By solving above differential equation

$T=20-15e^{Kt}$

When t=25 min T=10°C

$10=20-15e^{K\times 25}$

$K=\dfrac{1}{25}\ln\dfrac{2}{3}$

(a) When t=45 min

$T=20-15e^{45\times \dfrac{1}{25}\ln\dfrac{2}{3}}$

T=12.77°C

(b) When T=16°C

$16=20-15e^{t\times \dfrac{1}{25}\ln\dfrac{2}{3}}$

t=83.38 min