1 Answer

Terry Wei · Answer 1 · 2020-05-21T03:58:51+0000

Answer:

Density of water at
$T = 82^0 C , \rho = 983.308 kg/m^3$

Step-by-step explanation:

The relationship between density and temperature is shown below:

$\rho_1 = \rho_0 [ 1- \beta \Delta T ]$

Where,

$\rho_1$ is the density at temperature
T_1

$\rho_0$ is the density at temperature
T_0

$\beta$ is the coefficient of volume expansion

$\Delta T$ is the change in temperature which is:

$\Delta T = {T_1} -{T_0}$

Given,

T_0 = 4^0 C

$\rho_0 = 1.00* 10^3 kg/m^3$

T_1 = 82^0 C

$\Delta T = (82 -4) ^0 C =78 ^0 C$

$\rho_1 = ?$

Also,

$\beta for water = 0.000214 ^0C^(-1)$

So,

$\rho_1 is:$

$\rho_1 = 1.00* 10^3 kg/m^3[1 - 0.000214 ^0C^(-1) * 78^0 C ]$

$\rho_1 = 983.308 kg/m^3$

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