1320 btu`s
212 - 80 = 132 degrees temperature increase x 1 pound water = 132 BTU
6,520 Btus
One BTU is the energy required to raise one pound of water by one degrees. Therefore, your answer would be one half.
actually its 313.
313 Btu
A 100 pound propane tank has a capacity of 2,160,509 BTUs
There are 1078.9 BTUs.
1)This is a 5 part question. The first is realizing that 20F to 32F uses .5BTU per pound per degree. That means it takes 32-20=12*.5=6BTUs to get the ice to 32F. 2)Then you need to know the Latent Heat of Fusion for Ice which is 144BTUs (given). Lets assumes the ice changes from ice to water instantaneously at 32F. 3)Next we calculate the BTUs from 32F to 212F. Which is 1BTU per pound per degree F. 212-32=180 so it take 180BTUS. 4)Next we have to use the Latent Heat of Vaporization of water which will say instantaneously converts water to vapor. This takes 970BTUS (given). 5)Then we calculate the BTUS from 212F to 220F. Which is .5BTUs per pound per degree F which is 220-212=8*.5=4BTUs...... Finally add up all the BTUs and you get 6+144+180+970+4=1304BTUs.
212 - 80 = 132 degrees temperature increase x 1 pound water = 132 BTU
6,520 Btus
One BTU is the energy required to raise one pound of water by one degrees. Therefore, your answer would be one half.
1)This is a 5 part question. The first is realizing that 20F to 32F uses .5BTU per pound per degree. That means it takes 32-20=12*.5=6BTUs to get the ice to 32F. 2)Then you need to know the Latent Heat of Fusion for Ice which is 144BTUs (given). Lets assumes the ice changes from ice to water instantaneously at 32F. 3)Next we calculate the BTUs from 32F to 212F. Which is 1BTU per pound per degree F. 212-32=180 so it take 180BTUS. 4)Next we have to use the Latent Heat of Vaporization of water which will say instantaneously converts water to vapor. This takes 970BTUS (given). 5)Then we calculate the BTUS from 212F to 220F. Which is .5BTUs per pound per degree F which is 220-212=8*.5=4BTUs...... Finally add up all the BTUs and you get 6+144+180+970+4=1304BTUs.
There are no BTUs in an office water-cooler. But you can calculate how many BTUs are removed by the cooler. One BTU or British Thermal Unit is the amount of heat energy required to raise the temperature of one pound of water one degree Fahrenheit. There for when you remove one BTU you are lowering one pound of water one degree Fahrenheit. So if you know how many pounds of water you have and the temperature of the water you start with and the temperature of the water comming out of the cooler you can calculate how many BTUs the cooling unit of the water cooler has removed. BTU=Temp1 - Temp 2 X LB water
Well, you seem like a very smart kid. Let me ask you something: if you had one pound of naiveness in one hand and one pound of idiocy in the other, which one weighs more? Let me tell you a secret: Its the same!
144
1)This is a 5 part question. The first is realizing that 20F to 32F uses .5BTU per pound per degree. That means it takes 32-20=12*.5=6BTUs to get the ice to 32F. 2)Then you need to know the Latent Heat of Fusion for Ice which is 144BTUs (given). Lets assumes the ice changes from ice to water instantaneously at 32F. 3)Next we calculate the BTUs from 32F to 212F. Which is 1BTU per pound per degree F. 212-32=180 so it take 180BTUS. 4)Next we have to use the Latent Heat of Vaporization of water which will say instantaneously converts water to vapor. This takes 970BTUS (given). 5)Then we calculate the BTUS from 212F to 220F. Which is .5BTUs per pound per degree F which is 220-212=8*.5=4BTUs...... Finally add up all the BTUs and you get 6+144+180+970+4=1304BTUs.
1MW = 1000 KW 1KW = 860 Kcal/Hour 1 Kcal = 3.968 BTu.