Amount of energy required to heat a kg of water?

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Discussion Overview

The discussion revolves around the energy required to heat a kilogram of water from 50 to 70 degrees Celsius, considering the impact of heat loss to the surrounding atmosphere. Participants explore the complexities of heat transfer, including specific heat, insulation, and environmental factors.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant mentions the formula Q=mcΔT for calculating the energy needed to heat water, questioning if there is an equation that accounts for heat loss to the atmosphere.
  • Another participant suggests that the energy required depends on insulation against conduction losses, humidity, and the rate of heating, indicating that faster heating may reduce heat loss.
  • It is noted that the specific heat of air is much lower than that of water, which participants relate to everyday experiences.
  • Further contributions highlight the influence of convection currents and the geometry of the water container on heat transfer rates.
  • One participant discusses different heating methods, such as immersion heaters and microwaves, noting that these methods may minimize heat loss compared to heating water in a pot over a campfire.
  • Another participant points out that when using a gas stove, heat loss may be less significant due to the pot being heated directly by the flame.
  • A suggestion is made to consider conductive heat transfer between the water surface and air, referencing Newton's law of cooling.

Areas of Agreement / Disagreement

Participants express various viewpoints on the factors affecting heat loss and energy requirements, indicating that multiple competing views remain without a consensus on a single approach or solution.

Contextual Notes

Participants mention several variables that could affect the calculations, such as insulation quality, humidity, heating rate, and the specific heating method used, but do not resolve how these factors quantitatively influence the energy required.

tummbacoco
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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!
 
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tummbacoco said:
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.
 
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.
 
haruspex said:
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.
 
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.
 
...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.
 
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
 

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