How much energy is released or observed when 10 gm of steam at 100 degree Celsius turns to ice 0 degree Celsius
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Answer:
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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