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one gram of water at 373K is converted into steam at the
same temperature. the volume of 1 cubic cm of water bcums 1671cubic cm on
boiling. calculate change in internal energy if latent heat of vaporisation is
540 cal/g.
same temperature. the volume of 1 cubic cm of water bcums 1671cubic cm on
boiling. calculate change in internal energy if latent heat of vaporisation is
540 cal/g.
Verified Answer
one gram of water at 373K is converted into steam at thesame temperatu...
here we shall use the first law of thermodynamics
ΔQ = ΔU + ΔW
so, change in internal energy will be
ΔU = ΔQ - ΔW
here,
ΔQ = mL =1g x 540Cal/g = 540 Cal
ΔU = change in internal energy (to be calculated)
ΔW = change in work done
so,
ΔW = pΔV
where
ΔV = 1671cm3 - 1cm^3 = 1670cm^3 = 1670 x 10-6 m^3
p = hρg = 0.76 x 13600 x 9.8 ~ 101293 N/m^2
so,
ΔW = 101293 x 1670x10^-6 = 169.16J
or
ΔW = 169.16 / 4.2 = 40.3 Cal
thus, we have
ΔU = 540 Cal - 40.3 Cal
so,
ΔU = 499.7 Cal = 2098.74 J
This question is part of UPSC exam. View all JEE courses
Most Upvoted Answer
one gram of water at 373K is converted into steam at thesame temperatu...
Given Information:
- Mass of water = 1 gram
- Temperature of water = 373K
- Volume of water = 1 cm³
- Volume of steam = 1671 cm³
- Latent heat of vaporization = 540 cal/g
Calculating the change in internal energy:
Step 1: Calculate the heat absorbed by the water to reach its boiling point:
The specific heat capacity of water is 1 cal/g°C. Since the temperature change is from 373K to its boiling point, we can use the formula:
Q = mcΔT
Where:
Q = heat absorbed
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
Substituting the values:
Q = 1g * 1 cal/g°C * (373K - 273K)
Q = 100 cal
Step 2: Calculate the heat absorbed during vaporization:
The heat absorbed during vaporization can be calculated using the formula:
Q = mL
Where:
Q = heat absorbed
m = mass of water
L = latent heat of vaporization
Substituting the values:
Q = 1g * 540 cal/g
Q = 540 cal
Step 3: Calculate the change in internal energy:
The change in internal energy is the sum of the heat absorbed during heating and vaporization. Therefore:
ΔU = Q1 + Q2
Substituting the values:
ΔU = 100 cal + 540 cal
ΔU = 640 cal
Explanation:
When water is heated from 373K to its boiling point, it absorbs heat energy. This heat energy is used to increase the temperature of the water. Once the water reaches its boiling point, it undergoes a phase change from liquid to gas (steam). During this phase change, the water absorbs additional heat energy to overcome the intermolecular forces and convert into steam.
The change in internal energy is calculated by considering the heat absorbed during heating and vaporization. The heat absorbed during heating is calculated using the specific heat capacity of water and the temperature change. The heat absorbed during vaporization is calculated using the mass of water and the latent heat of vaporization.
In this case, 1 gram of water at 373K absorbs 100 cal of heat energy to reach its boiling point. Then, during vaporization, it absorbs an additional 540 cal of heat energy. Therefore, the total change in internal energy is 640 cal.
- Mass of water = 1 gram
- Temperature of water = 373K
- Volume of water = 1 cm³
- Volume of steam = 1671 cm³
- Latent heat of vaporization = 540 cal/g
Calculating the change in internal energy:
Step 1: Calculate the heat absorbed by the water to reach its boiling point:
The specific heat capacity of water is 1 cal/g°C. Since the temperature change is from 373K to its boiling point, we can use the formula:
Q = mcΔT
Where:
Q = heat absorbed
m = mass of water
c = specific heat capacity of water
ΔT = change in temperature
Substituting the values:
Q = 1g * 1 cal/g°C * (373K - 273K)
Q = 100 cal
Step 2: Calculate the heat absorbed during vaporization:
The heat absorbed during vaporization can be calculated using the formula:
Q = mL
Where:
Q = heat absorbed
m = mass of water
L = latent heat of vaporization
Substituting the values:
Q = 1g * 540 cal/g
Q = 540 cal
Step 3: Calculate the change in internal energy:
The change in internal energy is the sum of the heat absorbed during heating and vaporization. Therefore:
ΔU = Q1 + Q2
Substituting the values:
ΔU = 100 cal + 540 cal
ΔU = 640 cal
Explanation:
When water is heated from 373K to its boiling point, it absorbs heat energy. This heat energy is used to increase the temperature of the water. Once the water reaches its boiling point, it undergoes a phase change from liquid to gas (steam). During this phase change, the water absorbs additional heat energy to overcome the intermolecular forces and convert into steam.
The change in internal energy is calculated by considering the heat absorbed during heating and vaporization. The heat absorbed during heating is calculated using the specific heat capacity of water and the temperature change. The heat absorbed during vaporization is calculated using the mass of water and the latent heat of vaporization.
In this case, 1 gram of water at 373K absorbs 100 cal of heat energy to reach its boiling point. Then, during vaporization, it absorbs an additional 540 cal of heat energy. Therefore, the total change in internal energy is 640 cal.
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one gram of water at 373K is converted into steam at thesame temperature. the volume of 1 cubic cm of water bcums 1671cubic cm onboiling. calculate change in internal energy if latent heat of vaporisation is540 cal/g.
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one gram of water at 373K is converted into steam at thesame temperature. the volume of 1 cubic cm of water bcums 1671cubic cm onboiling. calculate change in internal energy if latent heat of vaporisation is540 cal/g. for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about one gram of water at 373K is converted into steam at thesame temperature. the volume of 1 cubic cm of water bcums 1671cubic cm onboiling. calculate change in internal energy if latent heat of vaporisation is540 cal/g. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for one gram of water at 373K is converted into steam at thesame temperature. the volume of 1 cubic cm of water bcums 1671cubic cm onboiling. calculate change in internal energy if latent heat of vaporisation is540 cal/g..
one gram of water at 373K is converted into steam at thesame temperature. the volume of 1 cubic cm of water bcums 1671cubic cm onboiling. calculate change in internal energy if latent heat of vaporisation is540 cal/g. for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about one gram of water at 373K is converted into steam at thesame temperature. the volume of 1 cubic cm of water bcums 1671cubic cm onboiling. calculate change in internal energy if latent heat of vaporisation is540 cal/g. covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for one gram of water at 373K is converted into steam at thesame temperature. the volume of 1 cubic cm of water bcums 1671cubic cm onboiling. calculate change in internal energy if latent heat of vaporisation is540 cal/g..
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