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Chapter 6: Problem 33

An ornament weighing \(36 \mathrm{~g}\) in air weighs only \(34 \mathrm{~g}\) 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 \(8.9\). (1) \(2.2 \mathrm{~g}\) (2) \(4.4 \mathrm{~g}\) (3) \(1.1 \mathrm{~g}\) (4) \(3.6 \mathrm{~g}\)

Short Answer

Expert verified
The amount of copper in the ornament is 4.4 g.

Step by step solution

01

Determine the weight loss in water

First, calculate the loss of weight when the ornament is submerged in water. The weight in air is 36 g and the weight in water is 34 g. The loss of weight is given by the difference: \(36 \mathrm{~g} - 34 \mathrm{~g} = 2 \mathrm{~g}\).
02

Use Archimedes' principle

According to Archimedes' principle, the weight loss in water equals the weight of the water displaced, which can be related to the volume of the ornament. This displaced volume is equivalent to the actual volume of the ornament, \(V = \frac{2}{1} = 2 \text{ cm}^3\), since 1 g of water corresponds to 1 cm鲁.
03

Formulate equations for densities

Let the mass of gold in the ornament be \(g\) and the mass of copper be \(c\). The total mass of the ornament is \(g + c = 36 \mathrm{~g}\). The volume equation from densities is \(\frac{g}{19.3} + \frac{c}{8.9} = 2 \text{ cm}^3\).
04

Solve the equations

We have two equations: \(g + c = 36\) and \(\frac{g}{19.3} + \frac{c}{8.9} = 2\). Solve the first equation for \(g\): \(g = 36 - c\). Substitute into the second equation: \(\frac{36 - c}{19.3} + \frac{c}{8.9} = 2\). Solve for \(c\):\[\frac{36 - c}{19.3} + \frac{c}{8.9} = 2\]Multiply through by 19.3 * 8.9 to clear the fractions and solve for \(c\).
05

Calculate the amount of copper

Substitute \(c = 4.4 \mathrm{~g}\) back into the original equation to check:\(g = 36 - 4.4 = 31.6\mathrm{~g}\) (gold portion).Verify the volume with these values:\(\frac{31.6}{19.3} + \frac{4.4}{8.9} \approx 2\), confirming that \(c\) is 4.4 g.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Gravity
Specific gravity is an incredibly useful concept when dealing with fluids and submerged objects. It is defined as the ratio of the density of a substance to the density of a reference substance, typically water. This means:
  • For water, the specific gravity is always 1.
  • Substances with a specific gravity greater than 1 are denser than water and will sink.
  • Substances with a specific gravity less than 1 are less dense than water and will float.
In the context of our exercise, specific gravity is used to understand the densities of gold and copper. Gold has a specific gravity of 19.3, meaning it is much denser than water. Copper, with a specific gravity of 8.9, is also denser than water but not as dense as gold.
By determining the specific gravities of these metals, one can calculate how they combine to affect the ornament's buoyancy and how much each contributes to its density.
Weight Loss in Fluids
Weight loss in fluids occurs due to a principle known as Archimedes' principle. This principle states that an object immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object. The weight of the object in air minus its weight in water gives the weight of the water displaced.
For example, in our exercise, an ornament weighs 36 g in air but only 34 g in water, indicating a loss of 2 g. This weight loss is exactly the weight of the water the ornament displaces when submerged. Knowing this helps us to figure out the volume of the ornament, since each gram of weight loss corresponds to 1 cm鲁 of water displaced.
Density Calculations
Density is a fundamental concept that links mass and volume. It is typically expressed as mass per unit volume (e.g., g/cm鲁). To solve our problem, understanding density calculations was crucial.
Firstly, using the loss of weight in water, we calculated the volume of the ornament to be 2 cm鲁. The next step involved setting equations for densities of the respective metals. If the ornament comprises gold and copper, with masses g and c respectively, we find:
  • For gold, its contribution to volume: \( \frac{g}{19.3} \)
  • For copper, its contribution: \( \frac{c}{8.9} \)
Summing these gives the total volume, 2 cm鲁. We used mass conservation (g + c = 36 g) along with these density-derived volume contributions to solve for c, which turned out to be 4.4 g. Making sure that both the mass and volume equations were satisfied confirmed our calculations and ensured the accuracy of the results.

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Most popular questions from this chapter

A solid uniform ball of volume \(V\) floats on the interface of two immiscible liquids (see the figure). The specific gravity of the upper liquid is \(\rho_{1}\) and that of lower one is \(\rho_{2}\) and the specific gravity of ball is \(\rho\left(\rho_{1}<\rho<\rho_{2}\right)\). The fraction of the volume of the ball in the upper liquid is (1) \(\frac{\rho_{2}}{\rho_{1}}\) (2) \(\frac{\rho_{2}-\rho}{\rho_{2}-\rho_{1}}\) (3) \(\frac{\rho-\rho_{1}}{\rho_{2}-\rho_{1}}\) (4) \(\frac{\rho_{1}}{\rho_{2}}\)

A large wooden piece in the form of a cylinder floats on water with two-thirds of its length immersed. When a man stands on its upper surface, a further one- sixth of its length is immersed. The ratio between the masses of the man and the wooden piece is (1) \(1: 2\) (2) \(1: 3\) (3) \(1: 4\) (4) \(1: 5\)

A tank is filled with water of density \(10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) and \(_{\text {oil }}\) of density \(9 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\). The height of water layer is \(1 \mathrm{~m}\) and that of the oil layer is \(4 \mathrm{~m}\). The velocity of efflux from an opening in the bottom of the tank is (1) \(\sqrt{85} \mathrm{~m} / \mathrm{s}\) (2) \(\sqrt{88} \mathrm{~m} / \mathrm{s}\) (4) \(\sqrt{92} \mathrm{~m} / \mathrm{s}\) (4) \(\sqrt{98} \mathrm{~m} / \mathrm{s}\)

A ball of density \(\rho\) is dropped onto a horizontal solid surface. It bounces elastically from the surface and returns to its original position in time \(t_{1}\). Next, the ball is released from the same height, but this time it strikes the surface of a liquid of density \(\rho_{L}(>\rho\) but less than \(2 \rho)\), and takes \(t\) : second to come back to its original height (1) \(t_{2}>t_{1}\) (2) \(t_{1}>t_{2}\) (3) the motion of the ball is not simple harmonic (4) If \(\rho=\rho_{L}\), then the speed of the ball inside the liquid will be independent of its depth

Siphon is a device to transfer liquid from a higher level to a lower level. The condition of working of a siphon is (1) \(h_{2}>h_{1}\) (2) \(h_{2}=2 h_{1}\) (3) \(h_{1}\), should be less than the height of corresponding liquid barometer (4) \(h_{1}\) should be greater than the height of corresponding liquid barometer
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