KVPY SA Exam 2013 Question Paper

Question : 68 of 80

Marks: +1, -0

A 10.0 W electrical heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of the water and the container rose by 3 K in 15 minutes. The container is then emptied, dried, and filled with 2 kg of an oil. It is now observed that the same heater raises the temperature of the container-oil system by 2 K in 20 minutes. Assuming no other heat losses in any of the processes, the specific heat capacity of the oil is

$2.5 × 10^3 \JK^{-1} \kg^{-1}$
$5.1 × 10^3 \JK^{-1} \kg^{-1}$
$3.0 × 10^3 \JK^{-1} \kg^{-1}$
$1.5 × 10^3 \JK^{-1} \kg^{-1}$

Validate

Solution:

\Pt = \m_\w\S_\w\Δ\T + \m_\c\s_\c\Δ\T

10 × 15 × 60 = 0.5 × 4200 × 3 +

\m_\c\s_\c

× 3
9000 = 6300 +

\m_\c\s_\c

\m_\c\s_\c

= 900 J/k.
Now, for oil
10 × 20 × 60 =

2 × \S_0 × 2 + 900 × 2

12000 0 1800

= 4 \S_0

\S_0=10200/4

= 2.51 × 10^3 \J\/\kg-\k

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