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Alexus [3.1K]
3 years ago
9

What 4 things affect how great the change in temperature for a substance is?

Physics
2 answers:
yawa3891 [41]3 years ago
7 0

Explanation:

The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved.

Fiesta28 [93]3 years ago
5 0

Answer: Usually, increasing the temperature increases the solubility of solids and liquids. Increasing the temperature always decreases the solubility of gases.

Explanation:

When you add a solute to a solvent, the kinetic energy of the solvent molecules overcomes the attractive forces among solute particles.

The solute particles leave the surface of the solid and move into the dissolved (aqueous) phase. In the image below the mass of grey (-) balls and green (+) balls represent a salt crystal. As the salt dissolves, the positive and negative ions are pulled apart and become surrounded by water molecules.

If we heat the solvent, the average kinetic energies of its molecules increases. Hence, the solvent is able to dislodge more particles from the surface of the solute.

Thus, increasing the temperature increases the solubilities of substances. For example, sugar and salt are more soluble in water at higher temperatures.

But, as the temperature increases, the solubility of a gas in a liquid decreases. As temperature increases, the average kinetic energy of the gas molecules increases.

As a result, the gas molecules dissolved in the liquid are more likely to escape to the gas phase and not return.

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A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w
spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

  1. directly proportional to its length i.eR\propto l
  2. inversely proportional to its cross section area i.eR\propto \frac{1}{A}

Therefore

R=\rho\frac{l}{A}

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2

R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }

\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }......(1)

Again for tungsten:

R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }

\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }........(2)

Given that R_1=R_2   and    l_1=l_2

Dividing the equation (1) and (2)

\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}

\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}   [since R_1=R_2   and    l_1=l_2]

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}

\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}

\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}

Therefore the ratio of diameter of the copper to that of the tungsten is

\sqrt{3} :\sqrt{10}

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