Question

At 1 atm, how much energy is required to heat 57.0g of H2O(s) at -24.0 ^degree C to H2O(g) at 147.0^degree C? Please answer in kJ & explain in detail

Answer 1

Q = mc∆T

Q = heat energy (Joules, J), m = mass of a substance (kg)

c = specific heat (units J/kg∙K), ∆ is a symbol meaning "the change in"

∆T = change in temperature (Kelvins, K)

Specific heat of ice = 2.05 j/g Deg Cel

Speicfic heat of water = 4.184 j/g Deg Cel

Heat of vaporization = 2259 J g¯¹

specific heat capacity for gaseous water (steam) = 2.02 J g¯¹K¯¹

Heat of vaporization = 6010 Joules / Moles

Q = 57 x 2.05 x 24 + 57 x 6010 /18 + 57 x 4.184 x 100 + 57 x 2259 + 57 x 2.02 x 47

Q = 2818.08 + 19031.66 + 23848.8 + 128763 + 5411.58

Q = 179873 Joules

179873 Joules of energy is required to heat 57.0g of H2O(s) at -24.0 ^degree C to H2O(g) at 147.0^degree C