A calorimeter of negligible heat capacity contains 100 cc of water at $40^{0}$ C. The water cools to $35^{0}$ C in 5 minutes. The water is now replaced by K-oil of equal volume at $40^{0}$ C. Find the time taken for the temperature to become $35^{0}$ C under similar conditions. Specific heat capacities of water and K-oil are 4200 J $k g^{- 1} K^{- 1}$ and 2100 J $k g^{- 1} K^{- 1}$ respectively. Density of K-oil = 800 kg $m^{- 3}$ .

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Solution

V = 100 CC, $Δ θ = 5^{0}$ C, T = 5 min
for water

$\frac{m s Δ θ}{d t} = \frac{K A}{l}$

for kerosene

$\frac{m s}{d t} = \frac{K A}{l} \frac{1}{Δ θ}$

$\Rightarrow \frac{100 \times 10^{-} 3 \times 800 \times 2100}{t} = \frac{K A}{l} . \frac{1}{Δ θ}$

= $\frac{100 \times 10^{- 3} \times 1000 \times 4200}{5}$

$\Rightarrow t = \frac{5 \times 800 \times 2100}{1000 \times 4200} = \frac{2000}{1000}$

= 2 min.

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