Question

How much energy is required to required to heat a 100. G sample of water from 0.0 degrees C to 100.0 degrees C. (water, c=4.18 J/g C, Hv=2260 J/g, Hf=334 J/g) *

Answer 1

41,800 J

Step-by-step explanation:

Here water is not to be converted from ice into water and then heated and then converted to vapors. Here water is to be heated only, so

Heat required= mcΔT

= 100 × 4.18 × (100-0)

= 41,800 J

Answer 2

`Q = 301200\,J`

Step-by-step explanation:

The energy required to heat water from its fussion point to its boiling point is the sum of latent and sensible heat. That is to say:

`Q = m\cdot (L_(f)+c_(p,w)\cdot \Delta T + L_(v))`

`Q = (100\,g)\cdot [334\,(J)/(g) +(4.18\,(J)/(kg\cdot ^(\textdegree)C) )\cdot (100\,^(\textdegree)C)+2260\,(J)/(g) ]`

`Q = 301200\,J`