Question

A liquid with a certain surface area takes 10 minutes to cool from 80° C to 70° C. The time taken by it to cool from 80° C to 60° C is [The surrounding temperature being 40° C] ____________.

Options

• 9 minute

• 13.4 minute

• 19.6 minute

• 23.3 minute

Answer 1

A liquid with a certain surface area takes 10 minutes to cool from 80° C to 70° C. The time taken by it to cool from 80° C to 60° C is 23.3 minute.

Explanation:

According to Newton's law of cooing

`(theta_1 - theta_2)/"t" = "k" [(theta _1 + theta_2)/2 - theta_"s"]`

`(80 -70)/10 = "k" [(80 + 70)/2 - 40]`

1 = 35 k       .....(i)

`80 _ 60/"t" = "k" [80 + 60/2 - 40]`

`20/"t" = 30 "k"`     ...(ii)

Dividing equation (i) by (ii),

`"t"/20 = 35/30`

`"t" = 35/30 xx 20 = 23.3  "minute"`