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Test Index
Thermodynamics Part 3
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Section:
Physics
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© examsnet.com
Question : 16 of 100
Marks:
+1
,
-0
An ice cube of dimensions
60
cm
×
50
cm
×
20
cm
60\,\text{cm} \times 50\,\text{cm} \times 20\,\text{cm}
60
cm
×
50
cm
×
20
cm
is placed in an insulation box of wall thickness
1
cm
1\,\text{cm}
1
cm
. The box keeping the ice cube at
0
∘
C
0^{\circ}\,\text{C}
0
∘
C
of temperature is brought to a room of temperature
4
0
∘
C
40^{\circ}\,\text{C}
4
0
∘
C
The rate of melting of ice is approximately :
(Latent heat of fusion of ice is
3.4
×
1
0
5
J
kg
−
1
3.4 \times 10^{5}\,\text{J}\,\text{kg}^{-1}
3.4
×
1
0
5
J
kg
−
1
and thermal conducting of insulation wall is
0.05
Wm
−
1
∘
C
−
1
)
0.05\,\text{Wm}^{-1}{}^{\circ}\text{C}^{-1})
0.05
Wm
−
1
∘
C
−
1
)
[26-Jul-2022-Shift-2]
61
×
1
0
−
3
kg
s
−
1
61 \times 10^{-3}\,\text{kg}\,\text{s}^{-1}
61
×
1
0
−
3
kg
s
−
1
61
×
1
0
−
5
kg
s
−
1
61 \times 10^{-5}\,\text{kg}\,\text{s}^{-1}
61
×
1
0
−
5
kg
s
−
1
208
kg
s
−
1
208\,\text{kg}\,\text{s}^{-1}
208
kg
s
−
1
30
×
1
0
−
5
kg
S
−
1
30 \times 10^{-5}\,\text{kg}\,\text{S}^{-1}
30
×
1
0
−
5
kg
S
−
1
Validate
Solution:
d
Q
d
t
=
K
A
Δ
T
ℓ
\;\frac{dQ}{dt} = \;\frac{KA \Delta T}{\ell}
d
t
d
Q
=
ℓ
K
A
Δ
T
A
=
2
(
0.6
×
0.5
+
0.5
×
0.2
+
0.2
×
0.6
)
A = 2(0.6 \times 0.5 + 0.5 \times 0.2 + 0.2 \times 0.6)
A
=
2
(
0.6
×
0.5
+
0.5
×
0.2
+
0.2
×
0.6
)
=
2
(
0.3
+
0.1
+
0.12
)
= 2(0.3 + 0.1 + 0.12)
=
2
(
0.3
+
0.1
+
0.12
)
=
2
(
0.4
+
0.12
)
= 2(0.4 + 0.12)
=
2
(
0.4
+
0.12
)
=
2
(
0.52
)
= 2(0.52)
=
2
(
0.52
)
=
1.04
m
2
= 1.04\,\text{m}^2
=
1.04
m
2
R
t
h
=
ℓ
K
A
⇒
1
×
1
0
−
2
0.05
×
1.04
⇒
1
0
−
2
0.052
R_{th} = \;\frac{\ell}{KA} \Rightarrow \;\frac{1 \times 10^{-2}}{0.05 \times 1.04} \Rightarrow \;\frac{10^{-2}}{0.052}
R
t
h
=
K
A
ℓ
⇒
0.05
×
1.04
1
×
1
0
−
2
⇒
0.052
1
0
−
2
d
Q
d
t
=
Δ
T
R
t
h
⇒
40
×
0.052
1
0
−
2
⇒
2.08
×
1
0
2
J
/
s
\;\frac{dQ}{dt} = \;\frac{\Delta T}{R_{th}} \Rightarrow \;\frac{40 \times 0.052}{10^{-2}} \Rightarrow 2.08 \times 10^{2}\,\text{J}/\text{s}
d
t
d
Q
=
R
t
h
Δ
T
⇒
1
0
−
2
40
×
0.052
⇒
2.08
×
1
0
2
J
/
s
2.08
×
1
0
2
=
m
×
3.4
×
1
0
5
2.08 \times 10^{2} = m \times 3.4 \times 10^{5}
2.08
×
1
0
2
=
m
×
3.4
×
1
0
5
m
=
2.08
3.4
×
1
0
3
⇒
0.61
×
1
0
−
3
kg
/
s
m = \;\frac{2.08}{3.4 \times 10^{3}} \Rightarrow 0.61 \times 10^{-3}\,\text{kg}/\text{s}
m
=
3.4
×
1
0
3
2.08
⇒
0.61
×
1
0
−
3
kg
/
s
=
61
×
1
0
−
5
Kg
/
s
= 61 \times 10^{-5}\,\text{Kg}/\text{s}
=
61
×
1
0
−
5
Kg
/
s
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