Answer:
The latent heat of vaporization of water is 2.4 kJ/g
Explanation:
The given readings are;
The first (mass) balance reading (of the water) in grams, mā = 581 g
The second (mass) balance reading (of the water) in grams, mā = 526 g
The first joulemeter reading in kilojoules (kJ), Qā = 195 kJ
The second joulemeter reading in kilojoules (kJ), Qā = 327 kJ
The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature
Based on the measurements, we have;
The latent heat of vaporization = ĪQ/Īm
ā“ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g
The latent heat of vaporization of water = 2.4 kJ/g