Question

Newton’s law of cooling states that for a cooling substance with initial temperature T0 , the temperature T(t) after t minutes can be modeled by the equation T(t)=Ts+(T0−Ts)e^−kt , where Ts is the surrounding temperature and k is the substance’s cooling rate.

A liquid substance is heated to 80°C . Upon being removed from the heat, it cools to 60°C in 15 min.

What is the substance’s cooling rate when the surrounding air temperature is 50°C ?

Round the answer to four decimal places.


a.) 0.0687

b.) 0.0732

c.) 0.0813

d.) 0.0872

Answer 1

Answer:

0.0916 is the right answer

Step-by-step explanation:

i took the test

Answer 2

Answer:

Answer is option b) 0.0732

Step-by-step explanation:

Given :

Initial Temperature  --  $\rm T_0 = 80^{\circ} C$

Final Temperature --  $\rm T = 60^{\circ} C$

Time, t = 15 min

Surrounding Temperature -- $\rm T_s = 50^{\circ} C$

Calculation :

Use formula,

$\rm T(t) = T_s + (T_0-T_s)e^-^k^t$

$60= 50+(80-50)e^1^5^t$

$10=30\;e^-^1^5^t$

$\rm ln(10) = ln(30) -15k\; ln(e)$

$\rm k = 0.073240819$

$\rm k = 0.0732 \;\;(upto\; four\; decimal \;places)$

Therefore, answer is option b) 0.0732

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