How much thermal energy is absorbed by 750 grams of water when its temperature increases from 15.4 C to 86.3°C?

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Answer 1 · 2016-11-28T09:46:39

222.27 kJ

#Q = m*c*DeltaT#

Where ,

Q = Thermal Energy
m = mass of the matter,
c = Specific heat
ΔT = Temperature difference [ Delta T]

So there is "-

m = 750 gm
c = 4.184 Joule
ΔT = 70.9 Degree C

#Q = 750*4.184*70.9=222248#

22248 Joules
or 222.48 kJ