An ornament weighting 36g in air, weighs only 34g in water. Assuming that some copper is mixed with gold to prepare the ornament, find the amount of copper in it. Specific gravity of gold is 19.3 and that of copper is 8.9.

Explanation

To solve the problem of finding the amount of copper mixed with gold in the ornament, we can follow these steps:

### Step 1: Understand the Problem
We know the weight of the ornament in air and in water. The weight in air is 36g, and the weight in water is 34g. The difference in weight gives us the buoyant force acting on the ornament when submerged in water.

### Step 2: Calculate the Buoyant Force
The buoyant force (FB) can be calculated as:
\[ FB = W_{air} - W_{water} \]
Substituting the values:
\[ FB = 36g - 34g = 2g \]

### Step 3: Relate Buoyant Force to Volume Displaced
The buoyant force is also equal to the weight of the water displaced by the ornament:
\[ FB = \text{Density of water} \times g \times V \]
Where \( V \) is the volume of the ornament. Since the density of water is \( 1g/cm^3 \), we can simplify this to:
\[ FB = V \times g \]
Thus, we have:
\[ 2g = V \times g \]
Cancelling \( g \) from both sides gives us:
\[ V = 2 \, cm^3 \]

### Step 4: Express Volume in Terms of Masses
The volume of the ornament can also be expressed in terms of the masses of gold and copper:
\[ V = \frac{m_{Cu}}{\rho_{Cu}} + \frac{m_{Au}}{\rho_{Au}} \]
Where:
- \( m_{Cu} \) = mass of copper (x grams)
- \( m_{Au} \) = mass of gold = \( 36g - x \)
- \( \rho_{Cu} \) = density of copper = \( \frac{8.9g/cm^3}{g/cm^3} \)
- \( \rho_{Au} \) = density of gold = \( \frac{19.3g/cm^3}{g/cm^3} \)

### Step 5: Substitute Values
Substituting the densities into the volume equation gives:
\[ 2 = \frac{x}{8.9} + \frac{36 - x}{19.3} \]

### Step 6: Solve for x
To solve for \( x \), multiply through by the least common multiple of the denominators (8.9 and 19.3):
\[ 2 \times 8.9 \times 19.3 = 19.3x + 8.9(36 - x) \]
Calculating the left side:
\[ 2 \times 8.9 \times 19.3 = 343.54 \]
Now expand the right side:
\[ 343.54 = 19.3x + 320.4 - 8.9x \]
Combine like terms:
\[ 343.54 = (19.3 - 8.9)x + 320.4 \]
\[ 343.54 - 320.4 = 10.4x \]
\[ 23.14 = 10.4x \]
Now, solve for \( x \):
\[ x = \frac{23.14}{10.4} \approx 2.22g \]

### Final Result
The amount of copper in the ornament is approximately **2.22 grams**.

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