What amount of energy ( in KJ) does it take for a 50.0g sample of H2O...

What amount of energy ( in KJ) does it take for a 50.0g sample of H2O (l) at 0 degreee celsius to be frozeen to -15.0 celsius? specific heat capacities ice, 2.03 j/g c, liquid 4.184 j/g c, steam 2.0 j/ g c, delta H vap is 40.7 Kj/mol and Delta H fus is 6.02 kj/mol

Expert Solution

Break the problem down into stages.

First stage

the water is being frozen and that there is NO temperature change. The molar heat of fusion value is used at the solid-liquid phase change.

q1 = (6.02 kJ / mol) (50 g / 18.0 g/mol) = 16.72 kJ

The second stage is to freeze the ice from 0° C to - 15° C.
Here we use the formula,
q2 = mc(delta T)
Where Q is the heat energy absorbed by the ice, m is the mass of the ice, c is the specific heat of ice, and (delta T) is the change in the ice's temperature.

q2 = 50g * 2.03J/g-°C * (-15+0)°C

= - 1522.5 J = - 1.5225 kJ : negative sign represents chilling.

Total amount of energy = q1 + q2 = 18.2425 kJ

queen_honey_blossom answered 2 years ago

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