calculate the heat required to convert 3 kg of ice at -12 deg c kept in calorimeter to steam at 100 deg c at atmospheric pressure
Answers
amount of heat required = 9.112752 × 10^6 J ≈ 9.1 × 10^6 J
mass of ice , m = 3kg
temperature of ice , Ti = -12°C
heat required to increase the temperature of ice upto 0°C , h1 = ms∆T
= 3kg × 0.5cal/g.°C × {0°C -(-12°C)}
= 3000g × 0.5 × 12
= 18000 cal
heat required to melt ice of mass 3kg, h2 = ml_f
= 3000 g × 80 cal/g
= 240000 cal
now increase the temperature of water to 100°C , h3 = ms'∆T'
= 3000g × 1 cal/g.°C × 100°C
= 300000 cal
heat required to convert water into steam, h4 = ml_v
= 3000g × 540 cal/g
= 1620000 cal
so, total heat required = h1 + h2 + h3 + h4
= 18000 + 240000 + 300000 + 1620000
= 2,178,000 cal
= 2178 Kcal
we know, 1KCal = 4184 Joule
so, heat = 2178 × 4184 = 9,112,752 J
= 9.112752 × 10^6 J
also read similar questions: Heat required to convert one gram of ice at 0oc into steam at 100o c is
https://brainly.in/question/5079887
note that the latent heat of phase change and the change and the specific heat capacity must be considered.
MASS OF ICE ...
https://brainly.in/question/60306
The mass of ice is:
m
=
3
k
g
m=3kg
The initial temperature of ice is:
T
1
=
−
5
∘
C
T1=−5∘C
The final temperature of steam is:
T
2
=
125
∘
C
T2=125∘C
The heat required by the ice to make steam is,
Q
=
m
c
i
c
e
(
0
−
(
−
5
∘
C
)
)
+
m
L
f
+
m
c
w
(
100
∘
C
−
0
∘
C
)
+
m
L
v
+
m
c
v
(
T
2
−
100
∘
C
)
Q=mcice(0−(−5∘C))+mLf+mcw(100∘C−0∘C)+mLv+mcv(T2−100∘C)
Here, the mass of ice is m, the specific heat of water is
c
w
cw, the specific heat of ice is
c
i
c
e
cice, the specific heat of steam is
c
v
cv, the latent heat of fusion is
L
f
Lf and latent heat of vaporization is
L
v
Lv.
Substitute the known values,
Q
=
3
k
g
(
2093
J
/
k
g
⋅
∘
C
)
(
0
−
(
−
5
∘
C
)
)
+
3
k
g
(
3.4
×
10
5
J
/
k
g
)
+
3
k
g
(
4186
J
/
k
g
⋅
∘
C
)
(
100
∘
C
−
0
∘
C
)
+
(
3
k
g
)
(
2.3
×
10
6
J
/
k
g
)
+
3
k
g
(
2009
J
/
k
g
⋅
∘
C
)
(
125
∘
C
−
100
∘
C
)
=
31395
J
+
1020000
J
+
1255800
J
+
6900000
J
+
150675
J
=
9357870
J
=
9.357
×
10
6
J
Q=3kg(2093J/kg⋅∘C)(0−(−5∘C))+3kg(3.4×105J/kg)+3kg(4186J/kg⋅∘C)(100∘C−0∘C)+(3kg)(2.3×106J/kg)+3kg(2009J/kg⋅∘C)(125∘C−100∘C)=31395J+1020000J+1255800J+6900000J+150675J=9357870J=9.357×106J
Thus, the energy required is
9.357
×
10
6
J
9.357×106J.