At a temperature of `0^@C` , the mass and volume of a fluid are 825 kg and `1.17 m^3` . The coefficient of volume expansion is `1.26 xx10^(-3)//""^@C` . What is the density of the fluid when the temperature has risen to `20.0^@C` ?

Options

A. `688 kg//m^3`
B. `562kg //m^3`
C. `705 kg//m^3`
D. `641 kg//m^3`

Answer

To find the density of the fluid when the temperature has risen to \(20.0^\circ C\), we will follow these steps:

### Step 1: Identify Given Values
- Initial temperature, \(T_1 = 0^\circ C\)
- Final temperature, \(T_2 = 20^\circ C\)
- Mass of the fluid, \(m = 825 \, \text{kg}\)
- Initial volume, \(V_1 = 1.17 \, \text{m}^3\)
- Coefficient of volume expansion, \(\gamma = 1.26 \times 10^{-3} \, \text{°C}^{-1}\)

### Step 2: Calculate the Change in Temperature
\[ \Delta T = T_2 - T_1 = 20^\circ C - 0^\circ C = 20^\circ C \]

### Step 3: Use the Volume Expansion Formula
The formula for volumetric expansion is given by:
\[ V_2 = V_1 \times (1 + \gamma \Delta T) \]
Substituting the known values:
\[ V_2 = 1.17 \, \text{m}^3 \times \left(1 + (1.26 \times 10^{-3} \, \text{°C}^{-1}) \times 20 \, \text{°C}\right) \]
Calculating the term inside the parentheses:
\[ 1 + (1.26 \times 10^{-3} \times 20) = 1 + 0.0252 = 1.0252 \]
Now substituting back to find \(V_2\):
\[ V_2 = 1.17 \, \text{m}^3 \times 1.0252 \approx 1.20 \, \text{m}^3 \]

### Step 4: Calculate the Density at \(20^\circ C\)
Density (\(\rho\)) is defined as mass per unit volume:
\[ \rho = \frac{m}{V_2} \]
Substituting the values:
\[ \rho = \frac{825 \, \text{kg}}{1.20 \, \text{m}^3} \approx 687.5 \, \text{kg/m}^3 \]

### Final Result
The density of the fluid when the temperature has risen to \(20.0^\circ C\) is approximately:
\[ \rho \approx 687.5 \, \text{kg/m}^3 \]