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
Verified Correct Answer: A.
Physical Principles & Numerical Solution
specific heat capacity of water = 4200Jkg\(^{-1}\)k\(^{-1}\), Specific heat of vaporization of water = 2.26 x 10\(^6\)Jkg\(^{-1}\), mass of water = 0.02kg, water vapour temp. = 100ºC, water at 97ºC
The amount of heat required = ML + Mc\Delta\(\theta\)
= 0.02 x 2.26 x 10\(^6\) + 0.02 x 4200 x (100 - 97)
= 45,200 + 252 = 45,452J.
Source: WAEC / JAMB / NECO Classroom Archive (Record #72080)