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 $= 800 k g m^{- 3}$

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Solution

$V = 100 c c$
$△ θ = 5^{0} C$
$t = 5 m i n$
for water
$\frac{m S △ θ}{d t} = \frac{K A}{l} △ θ$
$= \frac{100 \times 10^{- 3} 1000 \times 4200}{5} = \frac{K A}{l}$
for kerosene
$\frac{m a}{a t} = \frac{K A}{l}$
$\frac{100 \times 10^{- 3} \times 800 \times 2100}{t} = \frac{100 \times 10^{- 3} \times 1000 \times 4200}{5}$
$T = \frac{5 \times 800 \times 2100}{1000 \times 4200} = 2 m i n$

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V=100cc△θ=50Ct=5minfor water mS△θdt=KAl△θ=100×10−31000×42005=KAlfor kerosenemaat=KAl100×10−3×800×2100t=100×10−3×1000×42005T=5×800×21001000×4200=2min