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The Amazing World of Solutes, Solvents, and Solutions - Effect of temperature on density

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Density (ρ\rho) is defined as the mass (mm) of a substance per unit volume (VV). It is mathematically expressed as ρ=mV\rho = \frac{m}{V}.

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In a solution, when the temperature increases, the particles gain kinetic energy and move further apart. This phenomenon is known as thermal expansion.

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As the temperature of a solution increases, its volume (VV) increases while the mass (mm) remains constant. Consequently, the density (ρ\rho) of the solution decreases.

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The relationship between temperature (TT) and density (ρ\rho) is generally inverse: as T↑T \uparrow, ρ↓\rho \downarrow.

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Density is usually measured in units like g/cm3\text{g/cm}^3 or kg/m3\text{kg/m}^3.

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Anomalous Expansion of Water: Unlike most substances, water reaches its maximum density at 4∘C4^\circ \text{C}. Below this temperature, its density decreases as it turns into ice.

📐Formulae

ρ=mV\rho = \frac{m}{V}

Mass (m)=ρ×V\text{Mass (m)} = \rho \times V

Volume (V)=mρ\text{Volume (V)} = \frac{m}{\rho}

ρ∝1V\rho \propto \frac{1}{V}

💡Examples

Problem 1:

A sugar solution has a mass of 250 g250\text{ g} and occupies a volume of 200 mL200\text{ mL} at 25∘C25^\circ \text{C}. After heating, the volume expands to 250 mL250\text{ mL}. Calculate the change in density.

Solution:

Initial Density (ρ1\rho_1): ρ1=250 g200 mL=1.25 g/mL\rho_1 = \frac{250\text{ g}}{200\text{ mL}} = 1.25\text{ g/mL}

Final Density (ρ2\rho_2): ρ2=250 g250 mL=1.0 g/mL\rho_2 = \frac{250\text{ g}}{250\text{ mL}} = 1.0\text{ g/mL}

Change in Density: 1.25−1.000.25\begin{array}{r} 1.25 \\ - 1.00 \\ \hline 0.25 \end{array} Density decreased by 0.25 g/mL0.25\text{ g/mL}.

Explanation:

When the solution was heated, the mass remained 250 g250\text{ g}, but the increase in temperature caused the volume to increase from 200 mL200\text{ mL} to 250 mL250\text{ mL}, resulting in a lower density.

Problem 2:

If the density of a salt solution is 1.2 g/cm31.2\text{ g/cm}^3 and its volume is 50 cm350\text{ cm}^3, what is the mass of the solution? If the density drops to 1.0 g/cm31.0\text{ g/cm}^3 upon heating, what is the new volume?

Solution:

  1. To find Mass (mm): m=ρ×V=1.2 g/cm3×50 cm3=60 gm = \rho \times V = 1.2\text{ g/cm}^3 \times 50\text{ cm}^3 = 60\text{ g}

  2. To find New Volume (VnewV_{new}): Vnew=mρnew=60 g1.0 g/cm3=60 cm3V_{new} = \frac{m}{\rho_{new}} = \frac{60\text{ g}}{1.0\text{ g/cm}^3} = 60\text{ cm}^3

Explanation:

The mass of the solution stays constant at 60 g60\text{ g} during heating. As the density decreases to 1.0 g/cm31.0\text{ g/cm}^3, the volume must increase to 60 cm360\text{ cm}^3 to maintain the same mass.

Effect of temperature on density Class 8 Notes & Examples