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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?
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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
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Two containers a and b are connected by a conducting solid cylindrical...
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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?
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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?.
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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.
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