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

Answer:

41,800 J

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

Answer:

$Q = 301200\,J$

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:

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

$Q = (100\,g)\cdot [334\,\frac{J}{g} +(4.18\,\frac{J}{kg\cdot ^{\textdegree}C} )\cdot (100\,^{\textdegree}C)+2260\,\frac{J}{g} ]$

$Q = 301200\,J$