A certain freezing process requires that room temperature be lowered from $40^{\circ} C$ at the rate of $5^{\circ} C$ every hour. What will be the room temprature 10 hours after the process begins ?

Video Solution

Text Solution

Verified by Experts

Rate= $- 5^{0} \frac{c}{h}$
Initial T= $40^{0}$
After 10 hours = 40-(5*10)=40-50= $- 10^{0} C$ .

Updated on:21/07/2023

Class 7 MATHS INTEGERS

Related Playlists

BOOK - NCERTCHAPTER - INTEGERSEXERCISE - EXERCISE 1.4
7videos
INTEGERS
BOOK - NCERTCHAPTER - FRACTIONS AND DECIMALSEXERCISE - EXERCISE 2.6
5videos
FRACTIONS AND DECIMALS
BOOK - NCERTCHAPTER - LINES AND ANGLESEXERCISE - SOLVED EXAMPLES
1videos
LINES AND ANGLES

Similar Practice Problems

Question 1 - Select One
A certain freezing process requires that room temperature be lowered from $40^{\circ}$ C at the rate of $5^{\circ}$ C per hour. What will be the room temperature $12$ hours after the process begins?
A $- 21^{\circ} C$
B $- 20^{\circ} C$
C $- 16^{\circ} C$
D $- 18^{\circ} C$
Question 1 - Select One
A certain freezing process requires that room temperature be lowered from 50°C at the rate of 4°C every hour. What will be the room temperature 18 hours after the process begins ?
A $28^{\circ}$ C
B $- 28^{\circ}$ C
C $22^{\circ}$ C
D $- 22^{\circ}$ C
Question 1 - Select One
Water is filled in a vessel which is kept in a room at a temperature of $20^{\circ} C$ . The temperature of the water is slowly increased. When the temperature of water is $40^{\circ} C$ , it loses heat at the rate of 10cal/s. What will be the rate of loss of heat when the temperature of water is $80^{\circ} C$ ?
A20cal/s
B30cal/s
C40cal/s
D60cal/s
Question 1 - Select One
The count rate of a radioactive sample falls from $4.0 \times 10^{6} s^{- 1}$ to $1.0 \times 10^{6} s^{- 1}$ in 20 hours. What will be the count rate after 100 hours from beginning ?
A $3.91 \times 10^{3} s^{- 1}$
B $3.91 \times 10^{2} s^{- 1}$
C $3.91 \times 10^{4} s^{- 1}$
D $3.91 \times 10^{6} s^{- 1}$
Question 1 - Select One
Imagine a system, that can keep the room temperature within a narrow range between $20^{\circ} C$ to $25^{\circ} C$ . The system includes a heat engine operating with variable power P = 3KT, where K is a constant coefficient, depending upon the thermal insulation of the room, the area of the walls and the thickness of the walls. T is temperature of the room in temperature drops lower than $20^{\circ} C$ , the engine turns on, when the temperature increase over $25^{\circ}$ C, the engine turns off, room looses energy at a rate of $K (T - T_{0}), T_{0}$ is the outdoor temperature. The heat capacity of the room is C.
Given $(T_{0} = 10^{\circ} C, ln (\frac{3}{2}) = 0.4, ln (\frac{6}{5}) = 0.18, \frac{C}{K}$ =750 SI-unit)
Suppose at t = 0, the engine turns off, after how much time interval, again, the engine will turn on
A10 min
B5 min
C1.125 min
D2.25 min
Question 1 - Select One
Imagine a system, that can keep the room temperature within a narrow range between $20^{\circ} C$ to $25^{\circ} C$ . The system includes a heat engine operating with variable power P = 3KT, where K is a constant coefficient, depending upon the thermal insulation of the room, the area of the walls and the thickness of the walls. T is temperature of the room in temperature drops lower than $20^{\circ} C$ , the engine turns on, when the temperature increase over $25^{\circ}$ C, the engine turns off, room looses energy at a rate of $K (T - T_{0}), T_{0}$ is the outdoor temperature. The heat capacity of the room is C.
Given $(T_{0} = 10^{\circ} C, ln (\frac{3}{2}) = 0.4, ln (\frac{6}{5}) = 0.18, \frac{C}{K}$ =750 SI-unit)
Suppose at t = 0, the engine turns on, after how much time interval again, the engine will turn off
A10 min
B5 min
C1.125 min
D2.25 min
Question 1 - Select One
Liquid is filled in a vessel which is kept in a room with temperature 20°C. When the temperature of the liquid is 80°C, then it loses heat at the rate of 60 cal/sec. What will be the rate of loss of heat when the temperature of the liquid is 40°C ?
A $180 c a l / sec$
B $40 c a l / sec μ m$
C $1 μ / m$
D $1 m$
Question 1 - Select One
For a certain gaseous reaction rise of temperature from $25^{\circ} C$ to $35^{\circ} C$ doubles the rate of reaction. What is the value of activation energy :-
A $\frac{10}{2.30 R \times 298 \times 308}$
B $\frac{0.693 R \times 298 \times 308}{10}$
C $\frac{2.303 \times 10}{298 \times 308 R}$
D $\frac{0.693 R \times 10}{290 \times 308}$

A certain freezing process requires that room temperature be lowered from $40^{\circ} C$ at the rate of $5^{\circ} C$ every hour. What will be the room temprature 10 hours after the process begins ?

Rate= $- 5^{0} \frac{c}{h}$
Initial T= $40^{0}$
After 10 hours = 40-(5*10)=40-50= $- 10^{0} C$ .

Related Playlists

Similar Practice Problems

A certain freezing process requires that room temperature be lowered from $40^{\circ}$ C at the rate of $5^{\circ}$ C per hour. What will be the room temperature $12$ hours after the process begins?

A certain freezing process requires that room temperature be lowered from 50°C at the rate of 4°C every hour. What will be the room temperature 18 hours after the process begins ?

The count rate of a radioactive sample falls from $4.0 \times 10^{6} s^{- 1}$ to $1.0 \times 10^{6} s^{- 1}$ in 20 hours. What will be the count rate after 100 hours from beginning ?

Liquid is filled in a vessel which is kept in a room with temperature 20°C. When the temperature of the liquid is 80°C, then it loses heat at the rate of 60 cal/sec. What will be the rate of loss of heat when the temperature of the liquid is 40°C ?

For a certain gaseous reaction rise of temperature from $25^{\circ} C$ to $35^{\circ} C$ doubles the rate of reaction. What is the value of activation energy :-

Similar Questions

NCERT-INTEGERS-EXERCISE 1.3

A certain freezing process requires that room temperature be lowered from 40∘C at the rate of 5∘C every hour. What will be the room temprature 10 hours after the process begins ?

Rate=−50ch Initial T=400 After 10 hours = 40-(5*10)=40-50=−100C.

Related Playlists

Similar Practice Problems

A certain freezing process requires that room temperature be lowered from 40∘ C at the rate of 5∘C per hour. What will be the room temperature 12 hours after the process begins?

A certain freezing process requires that room temperature be lowered from 50°C at the rate of 4°C every hour. What will be the room temperature 18 hours after the process begins ?

Water is filled in a vessel which is kept in a room at a temperature of 20∘C. The temperature of the water is slowly increased. When the temperature of water is 40∘C, it loses heat at the rate of 10cal/s. What will be the rate of loss of heat when the temperature of water is 80∘C ?

The count rate of a radioactive sample falls from 4.0×106s−1 to 1.0×106s−1 in 20 hours. What will be the count rate after 100 hours from beginning ?

Liquid is filled in a vessel which is kept in a room with temperature 20°C. When the temperature of the liquid is 80°C, then it loses heat at the rate of 60 cal/sec. What will be the rate of loss of heat when the temperature of the liquid is 40°C ?

For a certain gaseous reaction rise of temperature from 25∘C to 35∘C doubles the rate of reaction. What is the value of activation energy :-

Similar Questions

NCERT-INTEGERS-EXERCISE 1.3

A certain freezing process requires that room temperature be lowered from $40^{\circ} C$ at the rate of $5^{\circ} C$ every hour. What will be the room temprature 10 hours after the process begins ?

Rate= $- 5^{0} \frac{c}{h}$
Initial T= $40^{0}$
After 10 hours = 40-(5*10)=40-50= $- 10^{0} C$ .

A certain freezing process requires that room temperature be lowered from $40^{\circ}$ C at the rate of $5^{\circ}$ C per hour. What will be the room temperature $12$ hours after the process begins?

The count rate of a radioactive sample falls from $4.0 \times 10^{6} s^{- 1}$ to $1.0 \times 10^{6} s^{- 1}$ in 20 hours. What will be the count rate after 100 hours from beginning ?

For a certain gaseous reaction rise of temperature from $25^{\circ} C$ to $35^{\circ} C$ doubles the rate of reaction. What is the value of activation energy :-