Calculating Latent Heat of Vaporisation: 3 kW Kettle, 2.0 kg Water @ 100oC

In summary: So in this case, the number would be displayed as 5.34 * 10^5 J, and 0.341 would be displayed as 5.34x10^3 J.
  • #1
luigihs
86
0
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..
 
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  • #2
luigihs said:
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..

Unit prefixes and powers of ten: read here
 
  • #3
gneill said:
Unit prefixes and powers of ten: read here

Cheers but I still struggling how can i get 3.38x10^6
 
  • #4
MJ = megaJoules = 106 Joules
 
  • #5
gneill said:
MJ = megaJoules = 106 Joules


Ok but that doesn't make sense because I have another problem with the same type of answer
E =1.5kg * 4190 J kg-1 K-1 x 85o = 5.34*10^5 J = 534 kJ. and there is no 10^5 in the decimal prefixes ...
 
  • #6
534 kJ = 534 x 103 J = 5.34 x 10^5 J

The reason it was specified as 534 kJ is because there's no 105 prefix.

Standard prefixes are for powers that are multiples of three (kilo and above).
 
  • #7
gneill said:
534 kJ = 534 x 103 J = 5.34 x 10^5 J

The reason it was specified as 534 kJ is because there's no 105 prefix.

Standard prefixes are for powers that are multiples of three (kilo and above).

So the answer is in Kj because is 10^5 and is between 10^5 and 10^3 ??
 
  • #8
luigihs said:
So the answer is in Kj because is 10^5 and is between 10^5 and 10^3 ??

Essentially, yes.

Sometimes you'll see values like 0.341 kJ or 1500 MW where the convention is "bent" a bit. Usually this is done to make the particular number have common units with other like-values for comparison (like entries in a table of values), or where one particular prefix would not handle all of the values to be shown.
 

Related to Calculating Latent Heat of Vaporisation: 3 kW Kettle, 2.0 kg Water @ 100oC

1. What is latent heat of vaporisation?

Latent heat of vaporisation is the amount of heat energy required to change a substance from liquid to gas at a constant temperature.

2. How is latent heat of vaporisation calculated?

The formula for calculating latent heat of vaporisation is Q = mL, where Q is the energy in joules, m is the mass of the substance in kilograms, and L is the latent heat of vaporisation for that substance.

3. What is the specific latent heat of vaporisation for water at 100°C?

The specific latent heat of vaporisation for water at 100°C is 2260 kJ/kg.

4. How does the power of the kettle affect the calculation of latent heat of vaporisation?

The power of the kettle, which is given in watts, can be converted to joules per second. This value is then used in the formula Q = mL to calculate the latent heat of vaporisation.

5. Can the latent heat of vaporisation of water vary at different temperatures?

Yes, the latent heat of vaporisation for water can vary at different temperatures. It generally increases as the temperature increases, with the highest value being at the boiling point of 100°C.

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