Question

12. Calculate the energy needed to change 9 g of ice at -20oc to steam at 110 oc.

Answer 1

Explanation:

Converting 100. g of ice at 0.00 °C to water vapour at 100.00 °C requires 301 kJ of energy.

There are three heats to consider:

q

1

= heat required to melt the ice to water at 0.00 °C.

q

2

= heat required to warm the water from 0.00 °C to 100.00 °C.

q

3

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

q

1

=

m

Δ

H

f

u

s

=

100

.

g

×

334

J

1

g

= 33 400 J

q

2

=

m

c

Δ

T

=

100

.

g

×

4.184

J

1

°

C

∙

g

×

100.00

°

C

= 41 800 J

q

3

=

m

Δ

H

v

a

p

=

100

.

g

×

2260

J

1

g

= 226 000 J

q

1

+

q

2

+

q

3

= 33 400 J + 41 800 J + 226 000 J = 301 000 J

= 301 kJ

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