If 200 grams of water is to be heated from 24.0oC to 100.0oC to make a cup of tea, how much heat must be added?

According to this formula:

Q = M*C*ΔT

when we have M (the mass of water) = 200 g

and C (specific heat capacity) of water = 4.18 J/gC

ΔT (the difference in temperature) = Tf - Ti

= 100 - 24

= 76°C

So by substitution:

Q = 200 g * 4.18 J/gC * 76 °C

= 63536 J

∴ the amount of heat which be added and absorbed to raise the temp from 24°C to 100°C is = 63536 J