JEE Exam > JEE Questions > Two containers a and b are connected by a con...
Start Learning for Free
Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7?
Most Upvoted Answer
Two containers a and b are connected by a conducting solid cylindrical...
Problem Statement:
Two containers A and B are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3. Thermal conductivity of rod is 693 watt/mol-k. The container A contains two moles of oxygen gas and the container B has four moles of helium gas. At time t=0 temperature difference of the containers is 50°C. After what time temperature difference between them will be 25°C transfer of heat takes place through rod only. Neglect radiation loss. Take R=8.3 J/mol-k and pi =22/7.
Solution:
Given,
- Length of rod = 242/7 cm
- Radius of rod = sqrt(8.3)
- Thermal conductivity of rod = 693 watt/mol-k
- Number of moles of oxygen gas in container A = 2
- Number of moles of helium gas in container B = 4
- Temperature difference at t=0 = 50°C
- Final temperature difference required = 25°C
- Gas constant R = 8.3 J/mol-k
- Pi = 22/7
We need to find out the time required for the temperature difference to reduce from 50°C to 25°C.
Step 1: Calculation of Heat Transfer
The heat transfer through the rod can be calculated using the formula:
q = (k*A/L) * (T1-T2)*t
where,
- q = heat transferred
- k = thermal conductivity of rod
- A = area of cross-section of rod
- L = length of rod
- T1 = initial temperature of container A
- T2 = initial temperature of container B
- t = time taken
Area of cross-section of rod can be calculated using:
A = pi*r^2
where,
- r = radius of rod
After substituting the given values, we get:
A = pi*(sqrt(8.3))^2 = 8.3*pi
q = (693*8.3*pi/(242/7)) * (50-0) * t
=> q = 1501.25*pi*t
Step 2: Calculation of Heat Capacity
The heat capacities of oxygen and helium can be calculated using:
Cp = 5/2 * R
where,
- Cp = heat capacity at constant pressure
- R = gas constant
After substituting the given values, we get:
Cp(O2) = 5/2 * 8.3 = 20.75 J/mol-k
Cp(He) = 5/2 * 8.3 = 20.75 J/mol-k
The total heat capacity of the gases in container A can be calculated using:
Cp_total(A) = n(O2) * Cp(O2)
where,
- n(O2) = number of moles of oxygen gas in container A
Similarly, the total heat capacity of the gases in container B can be calculated using:
Cp_total(B) = n(He) * Cp(He)
where,
- n(He) = number of moles of helium gas in container B
After substituting the given values, we get:
Cp_total(A) = 2 * 20.75 = 41.5 J/k
Cp_total
Two containers A and B are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3. Thermal conductivity of rod is 693 watt/mol-k. The container A contains two moles of oxygen gas and the container B has four moles of helium gas. At time t=0 temperature difference of the containers is 50°C. After what time temperature difference between them will be 25°C transfer of heat takes place through rod only. Neglect radiation loss. Take R=8.3 J/mol-k and pi =22/7.
Solution:
Given,
- Length of rod = 242/7 cm
- Radius of rod = sqrt(8.3)
- Thermal conductivity of rod = 693 watt/mol-k
- Number of moles of oxygen gas in container A = 2
- Number of moles of helium gas in container B = 4
- Temperature difference at t=0 = 50°C
- Final temperature difference required = 25°C
- Gas constant R = 8.3 J/mol-k
- Pi = 22/7
We need to find out the time required for the temperature difference to reduce from 50°C to 25°C.
Step 1: Calculation of Heat Transfer
The heat transfer through the rod can be calculated using the formula:
q = (k*A/L) * (T1-T2)*t
where,
- q = heat transferred
- k = thermal conductivity of rod
- A = area of cross-section of rod
- L = length of rod
- T1 = initial temperature of container A
- T2 = initial temperature of container B
- t = time taken
Area of cross-section of rod can be calculated using:
A = pi*r^2
where,
- r = radius of rod
After substituting the given values, we get:
A = pi*(sqrt(8.3))^2 = 8.3*pi
q = (693*8.3*pi/(242/7)) * (50-0) * t
=> q = 1501.25*pi*t
Step 2: Calculation of Heat Capacity
The heat capacities of oxygen and helium can be calculated using:
Cp = 5/2 * R
where,
- Cp = heat capacity at constant pressure
- R = gas constant
After substituting the given values, we get:
Cp(O2) = 5/2 * 8.3 = 20.75 J/mol-k
Cp(He) = 5/2 * 8.3 = 20.75 J/mol-k
The total heat capacity of the gases in container A can be calculated using:
Cp_total(A) = n(O2) * Cp(O2)
where,
- n(O2) = number of moles of oxygen gas in container A
Similarly, the total heat capacity of the gases in container B can be calculated using:
Cp_total(B) = n(He) * Cp(He)
where,
- n(He) = number of moles of helium gas in container B
After substituting the given values, we get:
Cp_total(A) = 2 * 20.75 = 41.5 J/k
Cp_total
Community Answer
Two containers a and b are connected by a conducting solid cylindrical...
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7?
Question Description
Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7?.
Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7?.
Solutions for Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? defined & explained in the simplest way possible. Besides giving the explanation of
Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7?, a detailed solution for Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? has been provided alongside types of Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? theory, EduRev gives you an
ample number of questions to practice Two containers a and b are connected by a conducting solid cylindrical rod of length 242/7 cm and (radius)^2=8.3 .thermal conductivity of rod is 693 watt/mol-k.the container a contains two moles of oxygen gas and the container b has four moles of helium gas .at time t=0 temprature difference of the containers is 50c after what time temprature difference between them will be 25c transfer of heat takes place through rod only.neglect radiation loss. take R=8.3 J/mol-k and pi =22/7? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup