At `39.2°F` , specific volume and density of water are respectively

Options

A. Minimum and maximum
B. Minimum and minimum
C. Maximum and minimum
D. Maximum and maximum

Answer

To solve the question regarding the specific volume and density of water at 39.2°F, we will follow these steps:

### Step 1: Convert the temperature to Celsius
To find the specific volume and density of water at 39.2°F, we first need to convert the temperature from Fahrenheit to Celsius.

The formula for conversion is:
\[ °C = \frac{5}{9} (°F - 32) \]

Substituting the given temperature:
\[ °C = \frac{5}{9} (39.2 - 32) = \frac{5}{9} \times 7.2 = 4°C \]

### Step 2: Understand the properties of water at 4°C
At 4°C (or 39.2°F), water exhibits a unique property known as anomalous expansion. This means that water reaches its maximum density at this temperature.

### Step 3: Find the density of water at 4°C
The density of water at 4°C is approximately:
\[ \rho = 1 \, \text{g/cm}^3 = 1000 \, \text{kg/m}^3 \]

### Step 4: Calculate the specific volume of water
Specific volume (\(v\)) is defined as the volume occupied by a unit mass of a substance. It is the inverse of density:
\[ v = \frac{1}{\rho} \]

Substituting the density we found:
\[ v = \frac{1}{1000 \, \text{kg/m}^3} = 0.001 \, \text{m}^3/\text{kg} = 1000 \, \text{cm}^3/\text{kg} \]

### Step 5: Summarize the results
At 39.2°F (or 4°C):
- The density of water is \(1 \, \text{g/cm}^3\) or \(1000 \, \text{kg/m}^3\).
- The specific volume of water is \(1000 \, \text{cm}^3/\text{kg}\) or \(0.001 \, \text{m}^3/\text{kg}\).

### Final Answer
- Specific Volume: \(1000 \, \text{cm}^3/\text{kg}\)
- Density: \(1 \, \text{g/cm}^3\)

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