Question

How many joules of heat are absorbed when 70.0 grams of water is completely vaporised at its boiling point?
(a) 23,352
(b) 7,000
(c) 15,813
(d) 158, 200

Answer 1

 the old heat of vaporization question, you have 70 g or 0.07kg water. At standard atmosphere the heat of vaporization is 2256.40 kJ/kg so at standard conditions it would take about 158 kJ to vaprize all the liquid.

Answer 2

Answer:

d) 158,200

Explanation:

Quantity of water = 70g or 0.07kg water. (Given)

At a standard atmosphere the heat of vaporization is = 2256.40 kJ/kg

Therefore, at standard conditions it would take about 158 kJ to vapourize all the liquid. Thus following the equation -

E = Quantity × heat

=  70 g E =70 × 2260

=158200

Thus, the joules of heat absorbed when 70.0 grams of water is completely vaporised at its boiling point are 158,200.