A piece of gold-alminium alloy weighs 49N. When suspended from a spring balance and submerged in water it weighs 39.2N. What is the weight of Gold in the alloy if the specific gravity of gold is 19.3 and that if aluminium is 2.5?
find the volume
49 N - 39.2 N = weight of water displaced (Says Archimedes) = 9.8 N water weight
water is about 1000 kg / m^3
so
9.8 N = mass of water * g
if g = about 9.8 m/s^2
then mass of water displaced = 1 kg
so water volume = metal volume = 1/1000 meter^3
so you have a rock weighing 49 N (mass = 49/9.8 = 5 kg) of volume 0.001 m^3
Au mass + Al mass = 5 kg
Au vol + Al vol = 0.001 m^3
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Au mass = 19.8 Au vol * (1000 kg/m^3) = 19,800 * Au vol
Al mass = 2.5 Al vol * (1000 kg/m^3) = 2,500 *Al vol
so
19,800 Au vol + 2,500 *Al vol = 5
1,000 Au vol + 1,000 Al vol = 1
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19,800* Au vol + 2,500 *Al vol = 5
2,500 * Au vol + 2,500 *Al vol = 2.5
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17,300 Au vol = 2.5
Au vol = 1.45 * 10 ^-4 meters^3
Au mass = 1.45*10^-4 * 19,800 = 2.86 kg
Au weight = 2.86*9.8 = 28 Newtons on earrth
To find the weight of gold in the alloy, we need to determine the weight of the volume of water displaced by the alloy. Here are the steps to calculate it:
Step 1: Find the weight of the water displaced by the alloy.
The weight of the water displaced can be calculated using the formula:
Weight of water displaced = Weight in air - Weight in water
Given:
Weight in air = 49N
Weight in water = 39.2N
Weight of water displaced = 49N - 39.2N = 9.8N
Step 2: Find the volume of water displaced by the alloy.
The volume of water displaced is equal to the volume of the alloy.
Step 3: Find the weight of the volume of gold in the alloy.
We know that the specific gravity is defined as the ratio of a substance's density to the density of a reference substance. In this case, the reference substance is water.
Given:
Specific gravity of gold = 19.3
The density of water is 1g/cm³ or 1000kg/m³.
The density of gold can be calculated by rearranging the formula as follows:
Density of gold = Specific gravity of gold x Density of water
Density of gold = 19.3 x 1000kg/m³ = 19300 kg/m³
Step 4: Calculate the volume of gold in the alloy.
The volume of gold can be determined using the formula:
Volume of gold = Weight of water displaced / Density of gold
Given:
Weight of water displaced = 9.8N
Density of gold = 19300 kg/m³
Using the formula, we can calculate the volume of gold in meters cubed (m³).
Volume of gold = 9.8N / 19300 kg/m³ = 0.000507m³
Step 5: Convert the volume of gold to grams.
To convert the volume to grams, we multiply the volume by the density of gold.
Given:
Density of gold = 19300 kg/m³
To convert to grams, we multiply by 1000.
0.000507m³ x 19300 kg/m³ x 1000 = 98.071 grams
Therefore, the weight of gold in the alloy is approximately 98.071 grams.
To find the weight of gold in the alloy, we need to understand the concept of specific gravity and use it to determine the weight of the gold and aluminum separately.
1. Start by understanding the concept of specific gravity:
- Specific gravity is the ratio of the density of a substance to the density of a reference substance. In this case, we are comparing the density of gold and aluminum to that of water.
- The specific gravity of a material is equal to its density divided by the density of water (SG = ρ_material / ρ_water).
- Since we know the specific gravity values of gold (19.3) and aluminum (2.5), we can use them to determine the density of gold and aluminum relative to water.
2. Determine the density of gold and aluminum relative to water:
- Multiply the specific gravity of each material by the density of water (ρ_water = 1000 kg/m³) to find the density of the material.
- Density of gold (ρ_gold) = Specific Gravity of Gold (SG_gold) x Density of Water (ρ_water)
Density of gold (ρ_gold) = 19.3 x 1000 kg/m³ = 19,300 kg/m³
- Density of aluminum (ρ_aluminum) = Specific Gravity of Aluminum (SG_aluminum) x Density of Water (ρ_water)
Density of aluminum (ρ_aluminum) = 2.5 x 1000 kg/m³ = 2500 kg/m³
3. Use the density values to find the weight of gold in the alloy:
- Let's assume the weight of the gold in the alloy is W_gold and the weight of the aluminum is W_aluminum.
- We can set up an equation based on the given information:
W_gold + W_aluminum = Weight of the alloy in air
- The weight of the alloy in air is given as 49 N.
W_gold + W_aluminum = 49 N (Equation 1)
- When the alloy is submerged in water, the buoyant force acting on the alloy reduces the weight. Thus, we have:
W_gold + W_aluminum - buoyant force = Weight of the alloy in water
- The weight of the alloy in water is given as 39.2 N.
W_gold + W_aluminum - buoyant force = 39.2 N (Equation 2)
- Substitute the density values of gold and aluminum into the equation for buoyant force:
buoyant force = density of water x volume of alloy
- The volume of the alloy is the same as the volume of the gold and aluminum combined, so we can rewrite the equation as:
buoyant force = density of water x volume of alloy = (ρ_water x V) (Equation 3)
- Now we have three equations (Equation 1, Equation 2, and Equation 3) with three unknowns (W_gold, W_aluminum, and V). We can solve them simultaneously to find the weight of gold in the alloy.
4. Solve the equations:
- Rearrange Equation 1 to solve for W_gold:
W_gold = 49 N - W_aluminum (Equation 4)
- Substitute Equation 4 into Equation 2:
49 N - W_aluminum + W_aluminum - (ρ_water x V) = 39.2 N
49 N - ρ_water x V = 39.2 N
- Simplify the equation:
ρ_water x V = 9.8 N (Equation 5)
- Rearrange Equation 5 to solve for V:
V = 9.8 N / ρ_water = 9.8 N / 1000 kg/m³ = 0.0098 m³
- Substitute V back into Equation 3 to find the buoyant force:
buoyant force = ρ_water x V = 1000 kg/m³ x 0.0098 m³ = 9.8 N
- Substitute the value of the buoyant force into Equation 2:
49 N - W_aluminum - 9.8 N = 39.2 N
39.2 N - W_aluminum = 39.2 N
- Solve for W_aluminum:
W_aluminum = 0 N
- Substitute W_aluminum into Equation 4 to find W_gold:
W_gold = 49 N - 0 N = 49 N
5. Calculate the weight of gold in the alloy:
- The weight of gold in the alloy is equal to the weight of gold alone, which is 49 N.
Therefore, the weight of gold in the alloy is 49 N.