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April 16, 2014

April 16, 2014

Posted by **ridhi** on Tuesday, May 29, 2012 at 3:16am.

- physics(heat) -
**Elena**, Tuesday, May 29, 2012 at 5:49amV/t =3L/min => m/t = 3 kg /min =3/60 =0.05 kg/s

ΔT = 77-27 = 50ºC

Specific heat of water c =4 180 J/kg•ºC.

Amount of heat used per second is

ΔQ/t =m•c• ΔT/t = 0.05•4180•50 =10450 J/s

Heat of combustion r = 4•10^7 J/kg

Rate of combustion of fuel is

10450/ 4•10^7 =2.61•10^-4 kg/s =

=0.0157 kg/min =15.7 g/min.

- physics(heat) -
**Anonymous**, Tuesday, May 29, 2012 at 12:33pmthanks :)

- physics(heat) -
**ridhi**, Friday, June 1, 2012 at 3:10pmthank you so much :)

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