Question

How much heat energy is required to heat 20.8 g of water from 39.5∘C to 187.2∘C? Answer in Kilojoules

Answer 1

To answer this question, we need to find the heat required in 3 steps to reach the final state at 187.2 ^oC.

the heat required to raise the temperature from 39.5 ^oC to 100 ^oC + heat required to vaporize + heat required to raise the steam temperature to 187.2^oC

To raise the temperature of the water from 39.5 ^oC to 100 ^oC
q₁ = m.Cp.ΔT
   = 20.8 g ( 4.186 j/g °C)(100 ^oC - 39.5 ^oC)
   = 5268 J
   = 5.268 KJ
Heat required to convert water at 100 degrees to steam at 100 degrees
q₂ = ΔH_vap. (no. of moles)
= 40.65 kJ/mol x (20.8 g/18 g/mol)
= 47 KJ
Heat required to raise the steam at 100^oC to steam at 187.2 ^oC.
q₃ = m.Cp.ΔT
   = 20.8 g x 1.864 J/(g·^oC) x (187.2 - 100)
   = 3381 J
   = 3.381 KJ

Therefore total heat required Q = q₁+ q₂ + q₃
                                = 5.268 KJ + 47 KJ + 3.381 KJ
                               = 55.65 KJ
Hence, the answer is: 55.65 KJ