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One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Then
- a)The work done by the vaporizing water is 169 J
- b)Heat added to the water to vaporize it is 2256 J
- c)Increase in internal energy is 2087 J
- d)None of these
Correct answer is option 'A,B,C'. Can you explain this answer?
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Verified Answer
One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a c...
(a) The work done by the vaporizing water is
W = p(V2 - V1)
= (1.013 x 105 Pa) (1671 x 10-6 m3 -1 x 10-6m3)
= 169 J
(b) The heat added to the water to vaporize it is
Q = mLv = (10-3 kg) (2.256 x 106 J/kg) = 2256 J
(c) From the first law of thermodynamics, the change in internal energy is
ΔU= Q — W = 2256 J - 169 J = 2087 J
W = p(V2 - V1)
= (1.013 x 105 Pa) (1671 x 10-6 m3 -1 x 10-6m3)
= 169 J
(b) The heat added to the water to vaporize it is
Q = mLv = (10-3 kg) (2.256 x 106 J/kg) = 2256 J
(c) From the first law of thermodynamics, the change in internal energy is
ΔU= Q — W = 2256 J - 169 J = 2087 J
Most Upvoted Answer
One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a c...
(a) The work done by the vaporizing water is
W = p(V2 - V1)
= (1.013 x 105 Pa) (1671 x 10-6 m3 -1 x 10-6m3)
= 169 J
(b) The heat added to the water to vaporize it is
Q = mLv = (10-3 kg) (2.256 x 106 J/kg) = 2256 J
(c) From the first law of thermodynamics, the change in internal energy is
ΔU= Q — W = 2256 J - 169 J = 2087 J
W = p(V2 - V1)
= (1.013 x 105 Pa) (1671 x 10-6 m3 -1 x 10-6m3)
= 169 J
(b) The heat added to the water to vaporize it is
Q = mLv = (10-3 kg) (2.256 x 106 J/kg) = 2256 J
(c) From the first law of thermodynamics, the change in internal energy is
ΔU= Q — W = 2256 J - 169 J = 2087 J
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Community Answer
One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a c...
The given information states that 1 gram of water, which is equivalent to 1 cm3, is boiled at a constant pressure of 1 atm. We are asked to determine the work done by the vaporizing water, the heat added to the water to vaporize it, and the increase in internal energy.
The work done by the vaporizing water is given by the equation:
Work = Pressure * Change in Volume
Given:
Pressure = 1 atm = 1.013 * 105 Pa
Change in Volume = 1671 cm3 - 1 cm3 = 1670 cm3
Substituting the values into the equation, we get:
Work = (1.013 * 105 Pa) * (1670 cm3)
= 1.688 * 108 J
Therefore, the work done by the vaporizing water is 169 J (option A).
The heat added to the water to vaporize it can be calculated using the equation:
Heat = Mass * Latent Heat of Vaporization
Given:
Mass of water = 1 gram = 0.001 kg
Latent Heat of Vaporization (Ly) = 2.256 x 106 J/kg
Substituting the values into the equation, we get:
Heat = (0.001 kg) * (2.256 x 106 J/kg)
= 2256 J
Therefore, the heat added to the water to vaporize it is 2256 J (option B).
The increase in internal energy can be calculated by subtracting the heat absorbed from the work done:
Increase in Internal Energy = Heat - Work
Substituting the values, we get:
Increase in Internal Energy = 2256 J - 1.688 * 108 J
= -1.685 * 108 J
Therefore, the increase in internal energy is -1.685 * 108 J, which is not one of the given options.
Hence, the correct answers are options A (work done by the vaporizing water) and B (heat added to the water to vaporize it). Option C (increase in internal energy) is incorrect.
The work done by the vaporizing water is given by the equation:
Work = Pressure * Change in Volume
Given:
Pressure = 1 atm = 1.013 * 105 Pa
Change in Volume = 1671 cm3 - 1 cm3 = 1670 cm3
Substituting the values into the equation, we get:
Work = (1.013 * 105 Pa) * (1670 cm3)
= 1.688 * 108 J
Therefore, the work done by the vaporizing water is 169 J (option A).
The heat added to the water to vaporize it can be calculated using the equation:
Heat = Mass * Latent Heat of Vaporization
Given:
Mass of water = 1 gram = 0.001 kg
Latent Heat of Vaporization (Ly) = 2.256 x 106 J/kg
Substituting the values into the equation, we get:
Heat = (0.001 kg) * (2.256 x 106 J/kg)
= 2256 J
Therefore, the heat added to the water to vaporize it is 2256 J (option B).
The increase in internal energy can be calculated by subtracting the heat absorbed from the work done:
Increase in Internal Energy = Heat - Work
Substituting the values, we get:
Increase in Internal Energy = 2256 J - 1.688 * 108 J
= -1.685 * 108 J
Therefore, the increase in internal energy is -1.685 * 108 J, which is not one of the given options.
Hence, the correct answers are options A (work done by the vaporizing water) and B (heat added to the water to vaporize it). Option C (increase in internal energy) is incorrect.
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One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer?
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One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer?.
One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer?.
Solutions for One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? in English & in Hindi are available as part of our courses for Physics.
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One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer?, a detailed solution for One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? has been provided alongside types of One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 * 105 Pa). The heat of vaporization at this pressure is Ly= 2.256 x 106 J/kg. Thena)The work done by the vaporizing water is 169 Jb)Heat added to the water to vaporize it is 2256 Jc)Increase in internal energy is 2087 Jd)None of theseCorrect answer is option 'A,B,C'. Can you explain this answer? theory, EduRev gives you an
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