The amount of heat energy that will be extracted when 0.02kg of water vapor condenses to water at 97ºC is [specific heat capacity of water = 4200Jkg\(^{-1}\)k\(^{-1}\), Specific heat of vaporization of water = 2.26 x 10\(^6\)Jkg\(^{-1}\)]
a
45,452J
b
45,200J
c
25,200J
d
252J
Explanation
Correct Option
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prestigiousemirate
1 year ago
Q=mL
there is no need for change in temperature as no further detail was given regarding the liquid
Q=0.02×(2.26×10^6)
𝑄=4.52×10^4
Q=4.52×10^4
𝑄=45,200 J
my humble opinion
