Question

If the temperature of the air is 30°c and the substance cools from 100°c to 70°c in 15 minutes ,find time when the temperature will be 40°c

Answer 1

After 5 minutes the temperature will be 40°C.

Step-by-step explanation:

The substance cools from 100°C to 70°C in 15 minutes.

So, the rate of cooling is $\frac{100 - 70}{15} = 2$ degrees per minute.

Now, we know that the rate of cooling is equal to the rate of heating.

Then the air will be heated by 2 degrees per minute.

So, the air will be heated from 30°C to 40°C i.e. by (40 - 30) = 10°C in $\frac{10}{2} = 5$ minutes. (Answer)

Answer 2

Given:

Temperature of the surrounding (T₀) = 30°C

To find:

Time taken for cooling from 70°C to 40°C

Explanation:

• According to Newtons law of cooling, "The rate at which an object cools from its initial temperature is directly proportional to the difference between the temperatures of the system and surroundings."

              Mathematically which is, $\frac{dQ}{dt}$ ∝ $(T_{2}- T_{1} )$

       If temperature of the surrounding T₀ is given, then

         Rate of fall of temperature will be, $\frac{dQ}{dt}$ ∝ $(\frac{(T_{1}+ T_{2}) }{2}-T_{0} )$

• Example of Newton's law of cooling is when you keep hot boiled milk in open for some time, it temperature keeps dropping.

Solution:

Step 1

In the first situation, we have

T₀ = 30°C

T₁  = 100°C  

T₂  = 70°C

t = 15 min

Substituting the given values in the equation, we get

$\frac{dQ}{15}=K(\frac{(100+70)}{2}-30 )$

Hence,

$dQ=K(825)$     $(i)$

Step 2

Now, in second situation, we have

T₀ = 30°C

T₁  = 70°C

T₂ = 40°C

t = ?

Substituting the given values in the equation, we get

$\frac{dQ}{dt}=K(\frac{(70+40)}{2}-30 )$

Hence,

$\frac{dQ}{dt}=K(25)$      $(ii)$

Substituting the value (i) in (ii), we get

$\frac{K(825)}{t}=K(25)$

t = 33 min

Final answer:

Hence, the substance will take more 33 min to cool from 70°C to 40°C.