Latent heat of vapourization of water is 537 cal what does it mean
Answers
Answer:
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Explanation:
Heat added or subtracted for a phase change = Latent heat x Mass
Q = L heat M
Where,
Q = heat (calories or joules)
L heat = latent heat (calories/gram or joules/gram)
M = mass (grams)
If liquid water at 100°C is changed into steam, the heat added (the latent heat of vaporization) is 540 calories for every gram of water. If steam at 100°C is changed into the water at 100° C, 540 calories for every gram of steam must be subtracted. If ice at 0°C is changed into liquid water at 0°C, the heat added (the latent heat of melting) is 80 calories for every gram of ice. If liquid water at 0°C changes into ice at 0°C, 80 calories for every gram of liquid water must be subtracted.
Latent heats can be very large. For example, the latent heat of vaporization of water is 540 cal/g and the latent heat of freezing of water is 80 cal/g. Therefore, changing a given quantity of water to steam requires 5.4 times as much heat as warming it from 0°C (+32°F) to 100°C (212°F), and melting ice requires as much heat as warming water from 20°C (68°F) to 100°C.