Answer:
Density of water at [tex] T = 82^0 C , \rho = 983.308 kg/m^3 [/tex]
Explanation:
The relationship between density and temperature is shown below:
[tex] \rho_1 = \rho_0 [ 1- \beta \Delta T ][/tex]
Where,
[tex] \rho_1[/tex] is the density at temperature [tex]T_1[/tex]
[tex] \rho_0[/tex] is the density at temperature [tex]T_0[/tex]
[tex] \beta [/tex] is the coefficient of volume expansion
[tex] \Delta T [/tex] is the change in temperature which is:
[tex] \Delta T = {T_1} -{T_0} [/tex]
Given,
[tex]T_0 = 4^0 C [/tex]
[tex] \rho_0 = 1.00\times 10^3 kg/m^3 [/tex]
[tex]T_1 = 82^0 C [/tex]
[tex] \Delta T = (82 -4) ^0 C =78 ^0 C [/tex]
[tex] \rho_1 = ? [/tex]
Also,
[tex] \beta for water = 0.000214 ^0C^{-1} [/tex]
So,
[tex] \rho_1 is: [/tex]
[tex] \rho_1 = 1.00\times 10^3 kg/m^3[1 - 0.000214 ^0C^{-1} \times 78^0 C ][/tex]
[tex] \rho_1 = 983.308 kg/m^3 [/tex]
