First slide

Specific heat

Question

A 10 W electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and container rises by 3K in 15 min. The container is then emptied, dried and filled with 2 kg of oil. The same heater now raises the temperature of container oil system by 2K in 20 min. Assume there is no heat loss in the process and the specific heat of water is 4200 J kg^-1K^-1 , the specific heat of oil in the same limit is equal to

Difficult

A

1.50 x 10³
B

2.55 x 10³
C

3.00 x 10³
D

2.10 x 10³

Solution

As we know, work done by an electric heater
$i.e. m_{1} c_{1} Δ t + m_{2} c_{2} Δ t = work done$
So, $m_{1} c_{1} Δ t + m_{2} c_{2} Δ t = P_{1} t_{1}$
where, $m_{1} = 0.5 kg, specific heat, c_{1} = 4200 {Jkg}^{- 1} K^{- 1}$
$Δ t = Δ t_{1} = Δ t_{2} = 3 K$
$P_{1} = 10 W$
$t_{1} = 15 \times 60 = 900 s$
c₂ = specific heat capacity of container.
$So, 0.5 \times 4200 \times (3 - 0) + m_{2} c_{2} \times (3 - 0) = 10 \times 15 \times 60$
$\Rightarrow 2100 \times 3 + m_{2} c_{2} \times 3 = 9000$
$\Rightarrow m_{2} c_{2} = \frac{9000 - 6300}{3} = 900$
Similarly, in the case of oil, $m_{1} c_{o} Δ t + m_{2} c_{2} Δ t = P_{2} t_{2}$
$where, c_{o} = specific heat capacity of oil, P_{2} = 10 W$
$2 \times c_{0} \times 2 + 900 \times 2 = 10 \times 20 \times 60$
$\Rightarrow 4 c_{0} + 1800 = 12000$
$\Rightarrow c_{o} = 2550 = 2.55 \times 10^{3} {Jkg}^{- 1} K^{- 1}$

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