How much energy would be required to heat 10.0 g of water at 0 degrees C to 100 degrees C, and change it into steam at 100 degrees C

Q = 4180 J

Explanation:

Given data:

Mass of water = 10.0 g

Initial temperature = 0°C

Final temperature = 100°C

Heat required = ?

Solution:

Specific heat capacity of water is 4.18 j/g.°C.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 100°C - 0°C

ΔT = 100°C

Q = 10.0 g ×4.18 j/g.°C × 100°C

Q = 4180 J

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"