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Question: The density of water at 4°C is 1000 kg \(m^{- 3}\) and at 100°C it is \(958.4kmm^{- 3}\) The coeffic...

The density of water at 4°C is 1000 kg m−3m^{- 3} and at 100°C it is 958.4kmm−3958.4kmm^{- 3} The coefficient of volume expansion of water is

A

4.5×10−3oC−14.5 \times 1{0^{- 3}}^{o}C^{- 1}

B

5.4×10−5oC−15.4 \times 1{0^{- 5}}^{o}C^{- 1}

C

4.5×10−4oC−14.5 \times 1{0^{- 4}}^{o}C^{- 1}

D

5.4×10−6oC−15.4 \times 1{0^{- 6}}^{o}C^{- 1}

Answer

4.5×10−4oC−14.5 \times 1{0^{- 4}}^{o}C^{- 1}

Explanation

Solution

As ρT2=ρT11+γΔT=ρT11+γ(T2−T1)\rho_{T_{2}} = \frac{\rho_{T_{1}}}{1 + \gamma\Delta T} = \frac{\rho_{T_{1}}}{1 + \gamma(T_{2} - T_{1})}

Here, T1=4∘C,T2=100∘CT_{1} = 4{^\circ}C,T_{2} = 100{^\circ}C

ρT1=1000kgm−3,ρT2=958.4kgm−3\rho_{T1} = 1000kgm^{- 3},\rho_{T2} = 958.4kgm^{- 3}

∴958.4=10001+γ(100−4)\therefore 958.4 = \frac{1000}{1 + \gamma(100 - 4)}

958.4=10001+96γor1+96γ=1000958.4958.4 = \frac{1000}{1 + 96\gamma}or1 + 96\gamma = \frac{1000}{958.4}

69γ=1000958.4−1=41.6958.469\gamma = \frac{1000}{958.4} - 1 = \frac{41.6}{958.4}

γ=4.16958.4×96=4.5×10−4∘C−1\gamma = \frac{4.16}{958.4 \times 96} = 4.5 \times 10^{- 4}{^\circ}C^{- 1}