The density of water at 20^oC is 0.998 , g/cm | vedclass

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(C) The relationship between density $\rho$ and temperature $t$ is given by $\rho_t = \rho_0 (1 - \gamma \Delta t)$,where $\gamma$ is the coefficient

Vedclass · Accepted Answer

$3 	imes 10^{-4}$

Vedclass · Answer

(C) The relationship between density $ho$ and temperature $t$ is given by $ho_t = ho_0 (1 - \gamma \Delta t)$,where $\gamma$ is the coefficient of cubical expansion.For a small change in temperature,the change in density is given by $ho_2 = ho_1 (1 - \gamma \Delta t)$.Rearranging this,we get $ho_2 - ho_1 = -ho_1 \gamma (t_2 - t_1)$.Taking the magnitude,$\gamma = \frac{ho_1 - ho_2}{ho_1 (t_2 - t_1)}$.Given $ho_1 = 0.998 \, g/cm^3$,$ho_2 = 0.992 \, g/cm^3$,$t_1 = 20^oC$,and $t_2 = 40^oC$.$\gamma = \frac{0.998 - 0.992}{0.998 	imes (40 - 20)} = \frac{0.006}{0.998 	imes 20} \approx \frac{0.006}{19.96} \approx 3.006 	imes 10^{-4} \, ^oC^{-1}$.Rounding to the nearest option,the value is $3 	imes 10^{-4} \, ^oC^{-1}$.

The density of water at $20^oC$ is $0.998 \, g/cm^3$ and at $40^oC$ is $0.992 \, g/cm^3$. The mean coefficient of cubical expansion (in per $^oC$) is:

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