Question

A body cools from 91℃ to 89℃ in 5 minutes when room temperature is 20℃. then it will cool from 71℃ to 69℃ in

Answer 1

Initial temp(T1) = 91

Final temp(T2)=89

surrounding temp(Tₐ)=20

Time taken to cool =5 minutes

The formula to be used will be, dT/dt=-k(T1+T2/2-Tₐ)

So from the initial given information we can find out the value of K

K=2/70

Now the next information is given as below

T1=71

T2=69

surrounding temp(Tₐ)=20

So the time taken can be calculated

2/dt= 2/70 ×(70-20)

∴dt = 7/5

      =1.2 minutes

Answer 2

As we know that,

Given:

A body cools from 91℃ to 89℃ in 5 minutes.

When the room temperature is 20℃.

Find out:

It will cool from 71℃ to 69℃ in?

Let us:

T1 = 91℃

T2 =89℃

t = 5 minute

Tₐ =20℃

Where;     T1 = initail temperature.

                T2 = final temperature.

                   t = time.

                 Tₐ = room temperature.

The formula;

dT/dt = -k(T1+T2/2-Tₐ)

so we can find out the value of K

K=2/70

Now,  

T1=71℃

T2=69℃

room temp(Tₐ)=20℃

Time taken can be calculated,

2/dt= 2/70 ×(70-20)

dt = 7/5

dt  =1.2 minutes.