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WBJEE 2014 Physics Solved Paper
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© examsnet.com
Question : 50
Total: 60
A 10 watt electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of waterand the container rises by 3° K in 15 minutes. The contain r is then emptied, dried and filled with 2 kg of oil. The sameheater now raises the temperature of container-oil system by 2°K in 20 minutes. Assuming that there is no heat lossin the process and the specific heat of water as 4200
J
k
g
–
1
K
–
, the specific heat of oil in the same unit is equal to
1.50 ×
10
3
2.55 ×
10
3
3.00 ×
10
3
5.10 ×
10
3
Validate
Solution:
(
1
2
×
4200
×
3
)
+
(
m
c
×
c
c
×
3
)
= 10 × 15 × 60 ... (1)
(
m
c
×
c
c
)
= 900. In case of oil
(
2
×
c
0
×
2
)
+
(
m
c
×
c
c
×
2
)
= (10 × 20 × 60) ,
C
0
+ (900 × 2) = 12000
(
C
0
)
= 2.55 ×
10
3
J
K
g
−
1
K
−
1
C
c
= Sp. heat capactiy of container
C
0
= Sp. heat capacity of oil
© examsnet.com
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