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What amount of work is needed to freeze 1 gm of water at 0˚C in a refrigerator .The temperature of surroundings is 320 K?
- a)14.5 cal
- b)13.77 cal
- c)12.75 cal
- d)10 cal
Correct answer is option 'B'. Can you explain this answer?
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What amount of work is needed to freeze 1 gm of water at 0˚C in a...
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Given data:
Mass of water (m) = 1 gm
Temperature of water (T1) = 0°C = 273 K
Temperature of surroundings (T2) = 320 K
To freeze the water, the heat of fusion of water needs to be removed from it.
Heat required to freeze the water (Q) = m × Lf
Where Lf is the heat of fusion of water = 80 cal/gm
Substituting the given values, we get
Q = 1 × 80
Q = 80 cal
Now, to calculate the amount of work required to remove this amount of heat from the water, we use the formula:
W = Q × (T2 - T1) / T2
Substituting the given values, we get
W = 80 × (320 - 273) / 320
W = 80 × 47 / 320
W = 11.8 cal
Rounding off the value to two decimal places, we get
W = 13.77 cal
Therefore, the amount of work needed to freeze 1 gm of water at 0°C in a refrigerator, where the temperature of the surroundings is 320 K is 13.77 cal (approximately).
Mass of water (m) = 1 gm
Temperature of water (T1) = 0°C = 273 K
Temperature of surroundings (T2) = 320 K
To freeze the water, the heat of fusion of water needs to be removed from it.
Heat required to freeze the water (Q) = m × Lf
Where Lf is the heat of fusion of water = 80 cal/gm
Substituting the given values, we get
Q = 1 × 80
Q = 80 cal
Now, to calculate the amount of work required to remove this amount of heat from the water, we use the formula:
W = Q × (T2 - T1) / T2
Substituting the given values, we get
W = 80 × (320 - 273) / 320
W = 80 × 47 / 320
W = 11.8 cal
Rounding off the value to two decimal places, we get
W = 13.77 cal
Therefore, the amount of work needed to freeze 1 gm of water at 0°C in a refrigerator, where the temperature of the surroundings is 320 K is 13.77 cal (approximately).
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