What is the change in the (liquid) water's temperature if the engine operates until 836000 J of Heat are added?
2007-02-07 14:06:35 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Q = m C (T2-T1)
where Q is the heat energy put into or taken out of the substance,
m is mass ,
C is the specific heat capacity, and
T2-T1 is the temperature differential
For liquid water C= 4.1813 Joules/(gram x deg Kelvin )
T2-T1=Q/(m C)
T2-T1= 836000 /(20 x1000x4.1813 )
T2-T1= 9.996 deg K or in this case C (10 degrees)
2007-02-08 01:46:05 · answer #1 · answered by Edward 7 · 0⤊ 0⤋
836000 X 0.0002389 / 20 = ~10 C. that's assuming that each and each body the added warmth is going purely into the water, none has been dissipated by technique of the radiator or via the engine block and none has been switched over to artwork.
2016-11-26 01:26:53 · answer #2 · answered by becher 3 · 0⤊ 0⤋