Home
Class 11
PHYSICS
A body cools in a surrounding of constan...

A body cools in a surrounding of constant temperature `30^@C` Its heat capacity is `2 J//^(@)C`. Initial temperature of cooling is valid. The body of mass 1 kg cools to `38^@C` in 10 min
When the body temperature has reached `38^@C`, it is heated again so that it reaches `40^@C` in 10 min. The heat required from a heater by the body is

A

`36^@C`

B

`36.4^@C`

C

`37^@C`

D

`37.5^@C`

Text Solution

Verified by Experts

The correct Answer is:
B
We have `theta-theta_s=(theta_0-theta_s)e^(-kt)`
Where `theta_0=`initial temperature of body `=40^@C`
`theta=`temperature of body after time t.
Since body cools from 40 to 38 in 10 min, we have
`38-30=(40-30)e^(-10k)` .(i)
Let after 10 min, the body temperature be `theta`
`theta-30^@=(38-30)e^(-10k)` .(ii)
`(Eq(i))/(Eq(ii))`gives `(8)/(theta-30)=(10)/(8)`,`theta-30=6.4`
`theta=36.4^@C`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • CALORIMETRY

    CENGAGE PHYSICS ENGLISH|Exercise Integer|8 Videos

  • CALORIMETRY

    CENGAGE PHYSICS ENGLISH|Exercise Solved Example|13 Videos

  • CALORIMETRY

    CENGAGE PHYSICS ENGLISH|Exercise Multiple Correct|25 Videos

  • BASIC MATHEMATICS

    CENGAGE PHYSICS ENGLISH|Exercise Exercise 2.6|20 Videos

  • CENTRE OF MASS

    CENGAGE PHYSICS ENGLISH|Exercise INTEGER_TYPE|1 Videos

Similar Questions

Explore conceptually related problems

A body having mass 2 Kg cools in a surrounding of constant temperature 30^(@)C . Its heat capacity is 2J//^(@)C . Initial temperature of the body is 40^(@)C . Assume Newton's law of cooling is valid. The body cools to 36^(@)C in 10 minutes. When the body temperature has reached 36^(@)C . it is heated again so that it reaches to 40^(@)C in 10 minutes. Assume that the rate of loss of heat at 38^(@)C is the average rate of loss for the given time . The total heat required from a heater by the body is :

A body cools in a surrounding of constant temperature 30^@C Its heat capacity is 2 J//^@C . Initial temeprature of cooling is valid. The body cools to 38^@C in 10 min The temperature of the body In .^@C denoted by theta . The veriation of theta versus time t is best denoted as

A body cools in a surrounding of constant temperature 30^@C Its heat capacity is 2 J//^(@)C . Initial temeprature of cooling is 40^@C and assuming newtons law of cooling is valid. The body cools to 38^@C in 10 min In further 10 min it will cools from 38^@C to _____

A body cools in a surrounding of constant temperature 30^(@)C . Its heat capacity is 2J//^(@)C . Initial temperature of the body is 40^(@)C . Assume Newton's law of cooling is valid. The body cools to 36^(@)C in 10 minutes. In further 10 minutes it will cool from 36^(@)C to :

A body is heated to a temperature 40^(@)C and kept in a chamber maintained at 20^@C . If the temperature of the body decreases to 36^@C in 2 minutues, then the time after which the temperature will further decrease by 4^@C , is

A body at temperature 40^(@)C is kept in a surrounding of constant temperature 20^(@)C . It is observed that its temperature falls to 35^(@)C in 10 minutes. Find how much more time will it taken for the body to attain a temperature of 30^(@)C .

A body cools in 7 min from 60^@C to 40^@C What will be its temperature after the next 7 min? The temperature of surroundings is 10^@C .

A body cools in 7 min from 60^@C to 40^@C What will be its temperature after the next 7 min? The temperature of surroundings is 10^@C .

If a body cools down from 80^(@) C to 60^(@) C in 10 min when the temperature of the surrounding of the is 30^(@) C . Then, the temperature of the body after next 10 min will be

A body cools from 50^@C to 40^@C in 5 mintues in surrounding temperature 20^@C . Find temperature of body in next 5 mintues.

Home
Profile