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Water (2510 g ) is heated until it just begins to boil. if the water absorbs 5.01×105 j of heat in the process, what was the initial temperature of the water?\

1 Answer

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E=energy=5.09x10^5J = 509KJ
M=mass=2250g=2.25Kg
C=specific heat capacity of water= 4.18KJ/Kg
ΔT= change in temp= ?
E=mcΔT
509=(2.25)x(4.18)xΔT
509=9.405ΔT
ΔT=509/9.405=54.1degrees
Initial temp = 100-54 = 46 degrees
Hope this helps :)
User Daxu
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