Calculate the heat energy required to vaporize 50g of water initially at 80ºC if the specific heat capacity of water is 4.23jg\(^{-1}\)k\(^{-1}\) (specific latent heat of vaporization of water is 2260jg\(^{-1}\)
530000J
23200J
17200J
130000J
Explanation
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Discussions (21)
Wrong answer.
The answer is not in the options though.
The formula used was correct but the value you used for change in temperature was wrong ( it should have been 100-80 =20•)
H = mc# +ml
= 50*4.23*20 + 50*2260
Wrong answer.
The answer is not in the options though.
The formula used was correct but the value you used for change in temperature was wrong ( it should have been 100-80 =20•)
H = mc# +ml
= 50*4.23*20 + 50*2260
= 117,230
For water to vaporize, temp. = 100°c
Hence, given initial temp. 80°c
We have: 100 - 80 = 20°c
H = mc∆T + mL = 4230 + 113000
= 117,230J
Option not given. 
These people ehn 🤦; the question says the INITIAL temperature is 80°C and since water was vaporized the temperature at which that occurs is 100°C which means that change in temperature= 100 - 80 = 20, but in their solution they used 80°C as their change in temperature. What the Hell!!!!!.😤
