If the molar heat of fusion of water is 6.01
1 answer:
Answer : The heat energy required to melt 2 kg of ice was, 667.7 kJ
Explanation :
First we have to calculate the moles of ice.
Now we have to calculate the heat energy.
As, heat energy required to melt 1 mole of ice = 6.01 kJ
So, heat energy required to melt 111.1 mole of ice = 111.1 × 6.01 kJ
= 667.7 kJ
Therefore, the heat energy required to melt 2 kg of ice was, 667.7 kJ
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Explanation:
The given data is as follows.
Temperature of dry bulb of air =
Dew point = = (10 + 273) K = 283 K
At the dew point temperature, the first drop of water condenses out of air and then,
Partial pressure of water vapor = vapor pressure of water at a given temperature
Using Antoine's equation we get the following.
= 0.17079
= 1.18624 kPa
As total pressure = atmospheric pressure = 760 mm Hg
= 101..325 kPa
The absolute humidity of inlet air =
= 0.00735 kg / kg dry air
Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg / kg dry air.
Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.
0.07 kg - 0.00735 kg
= 0.06265 kg for every 1 kg dry air
Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.
= 6.265 kg
Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.