- #1
luigihs
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A 3 kW kettle contains 2.0 kg of water at a temperature close to 100oC.
Latent heat of vaporisation for water: Lv=2256 (kJ kg^-1)
Q= Lv x mass
Ok I understand this problem because I now the answer but I don't understand the process.
Like my teacher wrote 2256x10^3 why he wrote 10^3 ?? I don't understand ! .. so he plugged into the equation and he get 2256x10^3 x 1.5kg = the first answer is 3.38x10^6 J But again I understand why he wrote 10^6? what's the point ... because If put in my calculator 2256 x 1.5 = 3384 ... , and the final answer is 3.38 MJ what happen to the 10^6 ?? why he convert to MJ I really confused help me please!
Thank you..
Latent heat of vaporisation for water: Lv=2256 (kJ kg^-1)
Q= Lv x mass
Ok I understand this problem because I now the answer but I don't understand the process.
Like my teacher wrote 2256x10^3 why he wrote 10^3 ?? I don't understand ! .. so he plugged into the equation and he get 2256x10^3 x 1.5kg = the first answer is 3.38x10^6 J But again I understand why he wrote 10^6? what's the point ... because If put in my calculator 2256 x 1.5 = 3384 ... , and the final answer is 3.38 MJ what happen to the 10^6 ?? why he convert to MJ I really confused help me please!
Thank you..