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To heat 1 pound of water by 1 degree Fahrenheit, you need 1 British Thermal Unit (BTU). This is based on the definition of a BTU, which is the amount of heat required to raise the temperature of one pound of water by one degree Fahrenheit at a constant pressure.

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Q: How many btu's to heat 1 POUND OF water 1 degree?
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How many BTUs does it take to heat one pound of water from 80 degree F to 212 degree F?

212 - 80 = 132 degrees temperature increase x 1 pound water = 132 BTU


How many BTUs will it take to change 1 gallon of water at 32 degrees to ice at 32 degrees?

You would need to remove approximately 1200 BTUs of heat to convert a gallon of water to ice. There are 8.34 lb in a gallon of water, which converting to lb-moles is 0.463. The latent heat of crystallization for water is -2583.4 BTU/lb-mole. Multiplying the two together and you get -1197 BTUs, which means you need to remove that amount of heat to convert the gallon of water to ice.


How many btu stored in 1 gal of 120 degree water?

one BTU is approximately the amount of energy needed to heat 1 pound of water 1 degree Fahrenheit. Here is a start ; now find out how much a pound of water weighs then work through it.


How much energy does it take to heat water from 32 deg f to 132 deg f?

1 BTU (british thermal unit) of heat energy will change the temperature of 1 pound of water by 1 degree Fahrenheit. If you are talking about 1 pound of water, since this is a change of 100 deg f, it would require 100 BTU. 2 pounds require 200 BTU. 3 pounds require 300 BTU. etc.


How much heat is required to change 0.2kg of ice at -5 degrees to water at 5 degrees?

Heat required for this transition is given as the the sum of three heatsheat required for heating the ice from -5 degree Celsius +latent heat(conversion of ice at zero degree to water at zero degrees)+heat required to heat the water from 0 to 5 degree CelsiusHeating of ice=m x s x delta T,where m is the mass ,s is the specific heat of ice=200x0.5x5=500calmelting of ice=mxlatent heat=200x80=16,000calHeating of water=m x s x delta T,where m is the mass ,s is the specific heat of water =200x1x5=1000calTotal heat required=500+16,000+1000=17,500 cal