Question

The heat of vaporization of water is 540 calories per gram. Based on your calculations in

question 13, how many calories would be required to raise the same mass of water and
convert it from water liquid (95 0C) to water gas (1000C)?
Show your work for partial credit.
Step 1: Water raises from 95°C to 1000C

Answer 1

To convert 100.0 g of water at 20.0 °C to steam at 100.0 °C requires 259.5 kJ of energy.

For this problem, there are only two heats to consider:

q

1

= heat required to warm the water from 20.0 °C to 100.0 °C.

q

2

= heat required to vapourize the water to steam at 100 °C.

q

1

=

m

c

Δ

T

=

100.0 g × 4.184 J

∘

C

−

1

g

−

1

×

80.0

∘

C

=

33 472 J

q

2

=

m

Δ

H

vap

=

100.0 g × 2260 J⋅g

−

1

=

226 000 J

q

1

+

q

2

=

( 33 472 + 226 000) J = 259 472 J = 259.5 kJ

(4 significant figures)