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Chapter 12: Problem 1

Density of ice is \(\rho\) and that of water is \(\sigma\). What will be the decrease in volume when a mass \(M\) of ice melts? (a) \(\frac{M}{\sigma-\rho}\) (b) \(\frac{a-\rho}{M}\) (c) \(M\left(\frac{1}{\rho}-\frac{1}{\sigma}\right)\) (d) \(\frac{1}{M}\left(\frac{1}{\rho}-\frac{1}{\sigma}\right)\)

Short Answer

Expert verified
The decrease in volume is \( M\left(\frac{1}{\rho}-\frac{1}{\sigma}\right) \) (option c).

Step by step solution

01

Understand the Problem

We need to find out the decrease in volume when a mass \( M \) of ice melts. We know the densities of ice \( \rho \) and water \( \sigma \). The volume decrease comes from the change in density between ice and water.
02

Calculate Initial Volume of Ice

The volume of ice before melting can be calculated using its density \( \rho \) and mass \( M \). The formula for volume \( V \) is \( V = \frac{M}{\rho} \).
03

Calculate Volume of Water After Melting

Once the ice melts, it becomes water. The volume of the resulting water can be calculated using the density of water \( \sigma \). The formula for volume is \( V = \frac{M}{\sigma} \).
04

Find Decrease in Volume

The decrease in volume when the ice melts is given by the difference in the two volumes calculated: \[ \text{Decrease in Volume} = \frac{M}{\rho} - \frac{M}{\sigma} = M \left( \frac{1}{\rho} - \frac{1}{\sigma} \right) \].
05

Match with Options

Compare our expression for the decrease in volume with the given options. The expression \( M \left( \frac{1}{\rho} - \frac{1}{\sigma} \right) \) matches option (c).

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Key Concepts

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

Volume Change
When a substance undergoes a phase change, such as from solid to liquid, its volume changes due to the differences in density. For ice melting into water, this means a reduction in volume occurs.
The initial volume of ice is larger than the volume of water produced because the density of ice is less than that of water.
This volume change is crucial for solving problems related to ice melting, as it directly impacts the calculations of how much volume decreases.
Melting Process
The melting process involves changing a solid, like ice, into a liquid, such as water, through the application of heat.
As ice absorbs heat, its molecules gain energy, breaking the rigid structure, and transitioning into the more fluid state of water.
While the mass remains constant during this process, the arrangement and spacing between molecules change.
  • Ice is organized in a crystalline lattice, resulting in lower density.
  • As it melts, molecules shift, packing more closely, resulting in higher density water.
Overall, this process leads to a decrease in volume, highlighted by the transition from ice to water.
Density of Ice
Density is a measure of mass per unit volume, and for ice, it is relatively low at approximately 0.917 grams per cubic centimeter (g/cm鲁).
This low density is due to the unique structure of ice, where water molecules form a lattice with open spaces that increase its volume.
These spaces make ice less dense than liquid water, which is why ice floats.
  • Understanding this helps to explain the volume change when ice melts.
  • Calculating the volume of ice before it melts relies on this density: \( V = \frac{M}{\rho} \).
With a knowledge of ice鈥檚 density, we can predict how melting will affect volume.
Density of Water
The density of liquid water is higher than that of ice, typically around 1 gram per cubic centimeter (g/cm鲁) at 4掳C.
This higher density means that for the same mass, water occupies less volume than ice.
The difference in density is key to understanding why there is a decrease in volume.
  • Water molecules are less orderly than in ice, allowing them to pack more closely.
  • The formula to find water鈥檚 volume after melting is \( V = \frac{M}{\sigma} \).
Recognizing the density difference allows us to determine how much the volume decreases as ice transitions to water during the melting process.

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Most popular questions from this chapter

The surface tension of a liquid at its boiling point (a) becomes zero (b) becomes infinity (c) is equal to the value at room temperature (d) is half to the value at the room temperature

A body of density \(\rho\) is dropped from rest at a height \(h\) into a lake of density \(\sigma\), where \(\sigma>\rho\). Neglecting all dissipative forces, calculate the maximum depth to which the body sinks before returning to float on the surface. (a) \(\frac{h}{a-\rho}\) (b) \(\frac{h \rho}{\sigma}\) (c) \(\frac{h \rho}{\sigma-\rho}\) (d) \(\frac{h \sigma}{\sigma-\rho}\)

A plane is in level flight at a constant apeed and each wing has an area of \(25 \mathrm{~m}^{2}\). During flight the speed of the air is \(216 \mathrm{kmh}^{-1}\) over the lower wing surface and \(252 \mathrm{kmh}^{-1}\) over the upper wing surface of each wing of aeroplane. Take density of air \(=1 \mathrm{kgm}^{-3}\) and \(g=10 \mathrm{~ms}^{-2}\) Pressure difference on each wing of aeroplane is (a) \(5 \mathrm{Nm}^{-2}\) (b) \(50 \mathrm{Nm}^{-2}\) (c) \(350 \mathrm{Nm}^{-2}\) (d) \(650 \mathrm{Nm}^{-2}\)

A bubble rises from bottom of a lake \(90 \mathrm{~m}\) deep. On reaching the surface, its volume becomes (take atmospheric pressure correspond upto \(10 \mathrm{~m}\) of water) [BVP Engg, 2006] (a) 18 times (b) 4 times (c) 8 times (d) 10 times

The surface area of air bubble increases four times when it rises from bottom to top of a water tank where the temperature is uniform. If the atmospheric pressure is \(10 \mathrm{~m}\) of water, the depth of the water in the tank is (a) \(30 \mathrm{~m}\) (b) \(40 \mathrm{~m}\) (c) \(70 \mathrm{~m}\) (d) \(80 \overline{\mathrm{m}}\)
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