(C) 100 g ice at 0°C is
converted into steam
at 100 °C. Find
total heat
> required (in kcal)
(L, = 80 cal/g,
sv = 1 cal/g°C,
L = 540 cal/g)
Answers
Answered by
1
Answer:
Explanation:
The amount of heat required is 36 kcal.
There are three separate heats involved in this problem:
q
1
= heat required to melt the ice to water at 0 °C
q
2
= heat required to warm the water from 0 °C to 100 °C
q
3
= heat required to convert the water to steam at 100 °C
q
=
q
1
+
q
2
+
q
3
=
m
Δ
fus
H
+
m
c
Δ
T
+
m
Δ
vap
H
q
1
=
m
Δ
fus
H
=
50
g
×
80
l
cal
⋅
g
-1
=
4 000 cal
q
2
=
m
c
Δ
T
=
50
g
×
1
l
cal
⋅
g
-1
°C
-1
×
100
°C
=
5 000 cal
q
3
=
m
Δ
vap
H
=
50
g
×
540
l
cal
⋅
g
-1
=
27 000 cal
q
=
q
1
+
q
2
+
q
3
=
4 000 cal + 5 000 cal + 27 000 cal
=
36 000 cal
=
36 kcal
Answered by
0
Answer:
Explanation:
Given: m = 150g
L = 540 cal/g
To find: Q = ?
Formula: Q = mL
Solution: Q = 150 × 540
= 81000 cal
Heat energy needed is 81000 cal.
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