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The heat required to convert 1 gm of water into steam under standard atmospheric pressure if the entropy of water at the boiling point is 0.31 and for 100% steam at the steam temperature is 1.74 _______ (in cal).
Correct answer is '533.4'. Can you explain this answer?
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The heat required to convert 1 gm of water into steam under standard a...
Q = 533.4 cals
ds = dQ/T
Q = T(S1 - S2)
Q = 373(1.74 - 0.31) = 533.4 cals
Q = 533.4 cals
ds = dQ/T
Q = T(S1 - S2)
Q = 373(1.74 - 0.31) = 533.4 cals
Q = 533.4 cals
Most Upvoted Answer
The heat required to convert 1 gm of water into steam under standard a...
Given:
- Mass of water (m) = 1 gm
- Entropy of water at boiling point (S1) = 0.31
- Entropy of steam at steam temperature (S2) = 1.74
To find:
- Heat required to convert 1 gm of water into steam under standard atmospheric pressure.
Solution:
Step 1: Calculation of change in entropy (ΔS):
The change in entropy during the phase transition can be calculated using the formula:
ΔS = S2 - S1
Substituting the given values:
ΔS = 1.74 - 0.31
ΔS = 1.43
Step 2: Calculation of heat required (Q):
The heat required for the phase transition can be calculated using the formula:
Q = m * ΔS
Substituting the given values:
Q = 1 * 1.43
Q = 1.43 cal
However, this value is in calories. To convert it into gram calories, we need to multiply it by 1000.
Q = 1.43 * 1000
Q = 1430 cal
Step 3: Conversion to standard units:
The answer is given in calories. To convert it into the standard unit of energy (Joule), we need to multiply it by 4.184.
Q = 1430 * 4.184
Q = 5977.12 J
Since the answer is given in cal, we convert it back to cal by dividing it by 4.184.
Q = 5977.12 / 4.184
Q = 1430.65 cal
Rounding off to the nearest tenth:
Q = 1430.7 cal
Therefore, the heat required to convert 1 gm of water into steam under standard atmospheric pressure is approximately 1430.7 cal.
Note: The given correct answer of 533.4 cal seems to be an error. The correct answer should be approximately 1430.7 cal as calculated above.
- Mass of water (m) = 1 gm
- Entropy of water at boiling point (S1) = 0.31
- Entropy of steam at steam temperature (S2) = 1.74
To find:
- Heat required to convert 1 gm of water into steam under standard atmospheric pressure.
Solution:
Step 1: Calculation of change in entropy (ΔS):
The change in entropy during the phase transition can be calculated using the formula:
ΔS = S2 - S1
Substituting the given values:
ΔS = 1.74 - 0.31
ΔS = 1.43
Step 2: Calculation of heat required (Q):
The heat required for the phase transition can be calculated using the formula:
Q = m * ΔS
Substituting the given values:
Q = 1 * 1.43
Q = 1.43 cal
However, this value is in calories. To convert it into gram calories, we need to multiply it by 1000.
Q = 1.43 * 1000
Q = 1430 cal
Step 3: Conversion to standard units:
The answer is given in calories. To convert it into the standard unit of energy (Joule), we need to multiply it by 4.184.
Q = 1430 * 4.184
Q = 5977.12 J
Since the answer is given in cal, we convert it back to cal by dividing it by 4.184.
Q = 5977.12 / 4.184
Q = 1430.65 cal
Rounding off to the nearest tenth:
Q = 1430.7 cal
Therefore, the heat required to convert 1 gm of water into steam under standard atmospheric pressure is approximately 1430.7 cal.
Note: The given correct answer of 533.4 cal seems to be an error. The correct answer should be approximately 1430.7 cal as calculated above.
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The heat required to convert 1 gm of water into steam under standard atmospheric pressure if the entropy of water at the boiling point is 0.31 and for 100% steam at the steam temperature is 1.74 _______ (in cal).Correct answer is '533.4'. Can you explain this answer?
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