J.R. S. answered 07/11/24
Ph.D. University Professor with 10+ years Tutoring Experience
Molar mass of nitromethane (CH3NO2) = 61.04 g / mole
moles nitromethane burned = 5.00 g x 1 mol / 61.04 g = 0.0819 moles
Heat generated from burning 0.0819 moles nitromethane = 0.0819 mol x 354.6 kJ/mol = 29.05 kJ (29,050 J)
To find the final temperature of the water, we will use q = mC∆T
q = heat = 29,050 J
m = mass of water = 250 g (assuming a density of 1 g / ml)
C = specific heat of water = 4.184 J/gº
∆T = change in temperature = ? (the temperature will rise from 20.5º to the final temp.)
Solving for ∆T we have...
∆T = q / (m)(C) = 29,050 J / (250 g)(4.184 J/gº)
∆T = 27.8º
Final temperature of the water = 20.5º + 27.8º = 48.3ºC