6 ms·

Is that tongue-in-cheek? Of course they can boil a kettle. It does take ages, though. Conventional kettles seemed to take perhaps twice as long to boil in CA

Is that tongue-in-cheek? Of course they can boil a kettle. It does take ages, though. Conventional kettles seemed to take perhaps twice as long to boil in CA than in UK. Boiling water at 230V,13A in the UK is going to be much faster than 120V,15A in the US.

wglb 14y ago
Um, no. Heating rate is a function of wattage, not voltage. EDIT: my error. I neglected to notice the amps shown and just picked up on the voltage.
- tacticus 14y ago
  W = V * I W1 = 230V * 13A = 2990 W2 = 120V * 15A = 1800 slight bit more power in the 220-240v at 10-13ish amp system than 120 @ 15
  - jsight 14y ago
    Don't most stoves in the US use 220? EDIT: oops, just realized that these are likely electric kettles. I was thinking of the stovetop variety.
    - tacticus 14y ago
      Yep these are separate plugin versions. the kettle i just bought on the weekend draws 2300W
- function_seven 14y ago
  I think he was implying power by listing the voltages and currents (230/13 vs. 120/15)
csense 14y ago
Assuming you're heating 300 mL of water. The specific heat capacity of water is 4.2 J / (g°C) [3]. If you're heating it from 20°C room temperature to 100°C boiling, that's a temperature increase of 80°C. The metric system is set up so that 1 mL of water weighs 1 g (give or take a little based on temperature, pressure, impurities, etc), i.e. the density of water is 1g / mL. Plugging in the numbers: (4.2 J/(g°C)) * (1 g/mL) * 300 mL * 80°C = 100800 J So approximately 100 kJ of energy is needed to boil 300 mL of water. One kW of power is a delivery of one kJ per second. How much power do we have available? In the UK, there's 230V * 13A = 2990 W ~ 3 kW. In the US, there's 120V * 15A = 1800 W = 1.8 kW. So in the UK it takes 100 kJ / 3 kW ~ 33 seconds. (One kW is 1 kJ / s.) In the US it takes 100 kJ / 1.8 kW ~ 56 seconds. These times are likely lower bounds. Assuming the voltage and current given by the parent are correct (or you measure them directly and substitute those numbers) and you measure out exactly 300 mL of distilled water, the biggest source of error is probably the fact that you're heating more than the water. You also have to heat the resistor in your kettle or electric stove, the thermally conducting, electrically insulating housing for the resistor, the kettle itself, and the surrounding air. And water molecules heat unevenly (my thermodynamics is a bit sketchy but I think this is called a Boltzmann distribution [2]?) so a few will evaporate before the water reaches a uniform 100°C, carrying off energy. My guess would be that the above effects matter less in the UK because the higher power means they would have less time to operate. I'm pretty sure that this analysis deals correctly with the fact that the current is AC [1]; if any experts on the mysteries of power engineering want to correct me, please do reply. [1] http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power http://en.wikipedia.org/wiki/Root_mean_square#Average_electr... [2] http://en.wikipedia.org/wiki/Boltzmann_distribution http://en.wikipedia.org/wiki/Boltzmann_distribution [3] http://en.wikipedia.org/wiki/Density_of_water#Heat_capacity_and_heats_of_vaporization_and_fusion http://en.wikipedia.org/wiki/Density_of_water#Heat_capacity_...

wglb 14y ago

Um, no. Heating rate is a function of wattage, not voltage. EDIT: my error. I neglected to notice the amps shown and just picked up on the voltage.

tacticus 14y ago
W = V * I W1 = 230V * 13A = 2990 W2 = 120V * 15A = 1800 slight bit more power in the 220-240v at 10-13ish amp system than 120 @ 15
- jsight 14y ago
  Don't most stoves in the US use 220? EDIT: oops, just realized that these are likely electric kettles. I was thinking of the stovetop variety.
  - tacticus 14y ago
    Yep these are separate plugin versions. the kettle i just bought on the weekend draws 2300W
function_seven 14y ago
I think he was implying power by listing the voltages and currents (230/13 vs. 120/15)

csense 14y ago

Assuming you're heating 300 mL of water. The specific heat capacity of water is 4.2 J / (g°C) [3]. If you're heating it from 20°C room temperature to 100°C boiling, that's a temperature increase of 80°C. The metric system is set up so that 1 mL of water weighs 1 g (give or take a little based on temperature, pressure, impurities, etc), i.e. the density of water is 1g / mL. Plugging in the numbers: (4.2 J/(g°C)) * (1 g/mL) * 300 mL * 80°C = 100800 J So approximately 100 kJ of energy is needed to boil 300 mL of water. One kW of power is a delivery of one kJ per second. How much power do we have available? In the UK, there's 230V * 13A = 2990 W ~ 3 kW. In the US, there's 120V * 15A = 1800 W = 1.8 kW. So in the UK it takes 100 kJ / 3 kW ~ 33 seconds. (One kW is 1 kJ / s.) In the US it takes 100 kJ / 1.8 kW ~ 56 seconds. These times are likely lower bounds. Assuming the voltage and current given by the parent are correct (or you measure them directly and substitute those numbers) and you measure out exactly 300 mL of distilled water, the biggest source of error is probably the fact that you're heating more than the water. You also have to heat the resistor in your kettle or electric stove, the thermally conducting, electrically insulating housing for the resistor, the kettle itself, and the surrounding air. And water molecules heat unevenly (my thermodynamics is a bit sketchy but I think this is called a Boltzmann distribution [2]?) so a few will evaporate before the water reaches a uniform 100°C, carrying off energy. My guess would be that the above effects matter less in the UK because the higher power means they would have less time to operate. I'm pretty sure that this analysis deals correctly with the fact that the current is AC [1]; if any experts on the mysteries of power engineering want to correct me, please do reply. [1] http://en.wikipedia.org/wiki/Root_mean_square#Average_electrical_power http://en.wikipedia.org/wiki/Root_mean_square#Average_electr... [2] http://en.wikipedia.org/wiki/Boltzmann_distribution http://en.wikipedia.org/wiki/Boltzmann_distribution [3] http://en.wikipedia.org/wiki/Density_of_water#Heat_capacity_and_heats_of_vaporization_and_fusion http://en.wikipedia.org/wiki/Density_of_water#Heat_capacity_...