Homework Help: physics

Posted by christy on Saturday, March 31, 2012 at 7:39pm.

In lab, Will burns a 0.6 g peanut beneath 48 g of water, which increases in temperature from 22 degree C to 55 degree C. The amount of heat absorbed by the water can be found with the equation Q =cm \Delta T, where Q is the amount of heat, c the specific heat of water, m the mass of water, and \Delta T the change in the water's temperature.
Assuming that 38 % of the heat released makes its way to the water, find the food value of the peanut.

physics - Damon, Saturday, March 31, 2012 at 7:50pm
heat into water = .38*heat out of peanut
48* Cw in cal/gm deg c * (55-22)/.38 = heat out of peanut in calories
divide by 1000 to get kilocalories(food calories)

that is the calories out of the peanut per 0.6 gram of peanut
if you want the kilocalories per gram divide by 0.6
physics - foziya, Monday, December 10, 2012 at 7:55pm
idk
physics - foziya, Monday, December 10, 2012 at 7:56pm
idk

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