Question

in the calibration of the calorimeter, an electrical resistance heater supplies 100.0 j of heat and a temperature increase of 0.850 degrees C is observed. Then 0.245 g of a particular fuel is burned in this same calorimeter, and the temperature increases by 5.23 degrees C. Calculate the energy density of this fuel, which is the amount of energy liberated per gram of fuel burned.

Answer 1

q =  Heater supplies / Temperature increase  = 100 / 0.850 = 118 J / C .

Heat = q * temperature increases = ( 118 * 5.23 ) = 615 j .

 The amount of energy liberated By the fuel = - 615 j .

Energy density =

  - The amount of energy liberated By the fuel / amount =

- ( -615 ) / 0.245 = 2510 j = 2.51 kg / g . Answer .