Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A body cools from 80 degree C to 50 degree C in 5 minutes. Calculate the time it takes to cool from 60 degree C to 30 degree C. The temperature of the surroundings is 20 degree C

Q. 11.20 A body cools from  80^{\circ}C to 50^{\circ}C  in 5 minutes. Calculate the time it takes to cool from 60^{\circ}C to  30^{\circ}C. The temperature of the surroundings is  20^{\circ}C.

Answers (1)

best_answer

Let a body initially be at temperature T1

Let its final Temperature be T2

Let the surrounding temperature be T0

Let the temperature change in time t.

According to Newton's Law of cooling

\\-\frac{dT}{dt}=K(T-T_{0})\\ \frac{dT}{T-T_{0}}=-Kdt\\ \int_{T_{1}}^{T_{2}}\frac{dT}{T-T_{0}}=-K\int_{0}^{t}dt\\ \left [ ln(T-T_{0}) \right ]_{T_{1}}^{T_{2}}=-K[t]_{0}^{t}\\ ln\left ( \frac{T_{2}-T_{0}}{T_{1}-T_{0}} \right )=-Kt

where K is a constant.

We have been given that the body cools from 80 oC to 50 oC in 5 minutes when the surrounding temperature is 20 oC.

T2 = 50 oC

T1 = 80 oC

T0 = 20 oC

t = 5 min = 300 s.

 \\ln\left ( \frac{50-20}{80-20} \right )=-K\times 300\\ K=\frac{ln(2)}{300}\\

For T1 = 60 oC and T2 = 30 oC we have

\\ln\left ( \frac{30-20}{60-20} \right )=-\frac{ln2}{300}t\\ t=\frac{ln(4)\times 300}{ln(2)} \\t=\frac{2ln(2)\times 300}{ln(2)}\\ t=600\ s\\ t= 10\ min

The body will take 10 minutes to cool from 60 oC to 30 oC at the surrounding temperature of 20 oC. 

Posted by

Sayak

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE Class 12 board exams 2025 to have 50% competency-based questions; no change in 10th
anam-khan

CBSE Class 12 board exams 2024 over; expected result date, trends of last 10 years

Latest Question