Hi, I believe that it lost 7000 cal. of energy. I believe that because it lost 26 degrees. Since a little cal. is how much energy it takes to raise 1 ml of water 1 degree. So I multiplied 28 by 250. That is how i got 7000 cal.
I believe it lost 7000 calories of energy. The specific heat of water is 1cal/gram*degree celsius and the mass was 250 grams. The temperature difference was 28 degrees celsius. 250g x 28 degrees x 1cal/gram x degrees celsius = 7000 calories.
I think that it lost 7000 calories. According to our last test, the formula for determining the energy loss is (mass)(change in temp.)(CP of Water). The mass us 250g x 28 degrees Celsius x 1 cal/(g)(degrees Celsius). The units of g and degrees Celsius cancel out so you are left with 250 x 28 x 1cal = 7000 cal.
6 comments:
Hi,
I believe that it lost 7000 cal. of energy. I believe that because it lost 26 degrees. Since a little cal. is how much energy it takes to raise 1 ml of water 1 degree. So I multiplied 28 by 250. That is how i got 7000 cal.
Thomas
Mr. Wilson, is that correct?
Didn't it lose 28 degrees not 26?
I believe it lost 7000 calories of energy. The specific heat of water is 1cal/gram*degree celsius and the mass was 250 grams. The temperature difference was 28 degrees celsius. 250g x 28 degrees x 1cal/gram x degrees celsius = 7000 calories.
I think that it lost 7000 calories. According to our last test, the formula for determining the energy loss is (mass)(change in temp.)(CP of Water). The mass us 250g x 28 degrees Celsius x 1 cal/(g)(degrees Celsius). The units of g and degrees Celsius cancel out so you are left with 250 x 28 x 1cal = 7000 cal.
Vivek basically summed it up...I agree with him, because his formula is correct and his calculations are correct.
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