Problem 64 - University Physics with Modern Physics

Question

A single ice cube with mass 9.70 g floats in a glass completely full of 420 \(\mathrm{cm}^{3}\) of water. You can ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 \(\mathrm{kg} / \mathrm{m}^{3} .\) What volume of salt water would the \(9.70-\mathrm{g}\) ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.

Step-by-Step Solution

Answer

(a) 10.58 cm³ (b) No overflow (c) 9.24 cm³ (d) Yes, 0.46 cm³ overflow.

Step 1: Convert Mass to Volume (Ice Cube)

To determine the volume of water displaced by the ice cube, use the density of ice. The density of ice is approximately 0.917 g/cm³. First, convert the mass of the ice cube into volume using the formula:\[\text{Volume of ice} = \frac{\text{Mass of ice}}{\text{Density of ice}} = \frac{9.70 \text{ g}}{0.917 \text{ g/cm}^3} = 10.58 \text{ cm}^3.\] The ice cube displaces 10.58 cm³ of water.

Step 2: Melting of Ice Cube in Water

When the ice cube melts, the volume of water it becomes is equal to the volume of water it originally displaced when it was in solid form. This is due to the principle of buoyancy and Archimedes' principle which states that the weight of the fluid displaced is equal to the weight of the object. Here, the ice cube weighed the same as the water it displaced (10.58 cm³). Therefore, no extra water is added to the glass, and there is no overflow.

Step 3: Displacing Salt Water

Now consider the ice cube floating in salt water with a density of 1050 kg/m³. First, convert this density to g/cm³ (1050 kg/m³ = 1.050 g/cm³). Use the formula to calculate the volume of salt water displaced by the ice cube:\[\text{Volume of salt water displaced} = \frac{\text{Mass of ice}}{\text{Density of salt water}} = \frac{9.70 \text{ g}}{1.050 \text{ g/cm}^3} = 9.24 \text{ cm}^3.\]This is because the ice cube needs to displace less volume as the salt water is denser than fresh water.

Step 4: Melting of Ice Cube in Salt Water

After melting in saltwater, the ice becomes freshwater with its original mass. The freshwater volume (9.70 g) is calculated as follows using the density of freshwater (1 g/cm³):\[\text{Volume of resulting water} = \frac{9.70 \text{ g}}{1 \text{ g/cm}^3} = 9.70 \text{ cm}^3.\]Since initially, it displaced 9.24 cm³ of saltwater, adding more (0.46 cm³ additional volume) to reach its melted state causes the water level to rise by this extra volume, potentially leading to an overflow.

Key Concepts

Archimedes' principle

Archimedes' principle is a fundamental concept in understanding buoyancy, which is the force exerted by a fluid that opposes the weight of an object immersed in it. According to this principle, when an object is partially or fully submerged in a fluid, the upward buoyant force experienced by the object is equal to the weight of the fluid it displaces.
This means that if an object is floating, the weight of the object is balanced by the weight of the fluid it displaces. In the given problem involving an ice cube floating in water, the cube displaces an amount of water equivalent to its own weight. Thus, the water displaced by the ice cube matches the volume of the ice cube as it floats, adhering to Archimedes' principle.

Density calculations

Density is a measure of how much mass is contained in a given volume. It is crucial for calculating how an object will behave in a fluid. The basic formula for density is \( \text{Density} = \frac{\text{Mass}}{\text{Volume}}\).
In the problem, to find out how much water the ice cube displaces, we use the density of ice, which is approximately 0.917 g/cm³. By knowing the mass of the ice cube (9.70 g) and using the formula for density, we determine the volume of ice and thus the volume of water displaced. The ice cube displaces a volume of water equal to \( \frac{9.70\ \text{g}}{0.917\ \text{g/cm}^3} = 10.58\ \text{cm}^3 \).
Doing density calculations helps us predict how much volume an object will take up in a fluid.

Ice melting and water displacement

When an ice cube melts in water, it turns from solid to liquid and becomes part of the liquid in the container. According to Archimedes' principle, the ice cube in solid form displaces a specific volume of water equivalent to the weight of the cube. When the ice melts, its volume as liquid water is exactly equal to the volume of water it originally displaced.
Thus, there is no change in the overall water level, and no overflow occurs. This phenomenon occurs because the density of water in liquid form compensates for the difference in volume between the solid and liquid states of the ice, ensuring consistent displacement before and after melting.

Saltwater vs freshwater buoyancy

The density of saltwater is typically higher than that of freshwater. In our exercise, the saltwater has a density of 1.050 g/cm³ compared to freshwater's 1 g/cm³. As density increases, a smaller volume of fluid needs to be displaced to counterbalance the weight of a floating object.
For the ice cube floating in saltwater, it displaces less volume because the saltwater is denser. The volume displaced in saltwater is \( \frac{9.70\ \text{g}}{1.050\ \text{g/cm}^3} = 9.24\ \text{cm}^3 \).
However, when the ice melts, it turns into freshwater, requiring \(9.70\ \text{cm}^3\) for the same mass. Since the initial displaced volume was less (9.24 cm³), adding the melted ice volume (9.70 cm³) causes an overflow of \(0.46\ \text{cm}^3\). This highlights the different buoyant effects between freshwater and saltwater.