Question

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?

Answer 1

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