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KEAM 2017 Physics and Chemistry Paper
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© examsnet.com
Question : 43
Total: 120
Initially a beaker had 100g of water at temperature 90°C. Later another 600g of water at temperature 20°C was poured into the beaker. The temperature, T, of the water after mixing is
20°C
30°C
45°C
55°C
90°C
Validate
Solution:
Given
Mass of water at
90
∘
C
=
100
g
m
=
100
×
10
−
3
k
g
Mass of water at
20
∘
C
=
600
g
m
=
600
×
10
−
3
k
g
From calorimetery
m
1
s
1
t
1
+
m
2
s
2
t
2
=
(
m
1
+
m
2
)
s
.
T
∵
s
1
t
1
+
s
2
t
2
=
s
t
[where,
T
is temperature of mixture].
100
×
10
−
3
×
1
×
90
+
600
×
10
−
3
×
1
×
20
=
(
100
+
600
)
×
10
−
3
×
1
×
T
T
=
100
×
10
−
3
×
90
+
600
×
10
−
3
×
20
700
×
10
−
3
=
(
9000
+
12000
)
×
10
−
3
700
×
10
−
3
=
21000
700
=
30
∘
C
© examsnet.com
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