Amount of energy required to heat a kg of water?

[tummbacoco]

So recently I've become familiar with concepts like specific heat, and we use a formula Q=mcΔT to calculate the Joules necessary to heat up say a kg of water from 50 to 70 degrees celsius. Now I was wondering if there is a equation that incorporates the amount of heat lost to the atmosphere, because you'd have to constantly add more and more energy just to fight off the cooling effect of the air around you. How much energy would it take to heat up a kg of water from 50 to 70 degrees celsius with a surrounding temp of 20 degrees C. Thanks!

 

[haruspex]

So recently I've become familiar with concepts like specific heat, and we use a formula Q=mcΔT to calculate the Joules necessary to heat up say a kg of water from 50 to 70 degrees celsius. Now I was wondering if there is a equation that incorporates the amount of heat lost to the atmosphere, because you'd have to constantly add more and more energy just to fight off the cooling effect of the air around you. How much energy would it take to heat up a kg of water from 50 to 70 degrees celsius with a surrounding temp of 20 degrees C. Thanks!

It depends on how well insulated the water is against conduction losses, the humidity of the air, and the rate of heating. The faster you can heat it, the lower the loss.

 

[Vanadium 50]

It's the same formula but a more complex system: you need to also consider the heat lost to (and change in temperature of) the air.

Note that the specific heat of air is much, much less than that of, say, water. This should agree with your everyday experience.

 

[Chestermiller]

It depends on how well insulated the water is against conduction losses, the humidity of the air, and the rate of heating. The faster you can heat it, the lower the loss.

It also depends on the convection currents in the room air and the geometry of the water container, which affect the rate of heat transfer to the air.

 

[256bits]

And I suppose the mode of heating.
An immersion heater would transfer all the energy to the water heating it up.
Heat loss would then only be that from the difference in temperature of the water and the air and all that that would encompass.
Utilizing a microwave oven would be much the same.
Heating the water in a pot from an external source leads to some of the energy being lost to raising the temperature of the air, and not at all the water. An extreme case would be heating the pot of water over a campfire. Probably what vanadium is referring to.

 

[russ_watters]

...though if we are talking about a pot on a gas stove, heat loss matters surprisingly little because the pot is bathed in the heat from the flame.

 

[weiwei]

You need to consider the conductive heat transfer between the water surface and air. Use q=c*(T_water - T_air). search Newton's law of cooling