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Alenkasestr [34]

3 years ago

10

Calculate the heat energy released when 13.3 g of liquid mercury at 25.00 C is converted to solid mercury at its melting point.C

onstants for mercury at 1 atmheat capacity of Hg(l) 28.0 J/(mol K)melting point 234.32 Kenthalphy of fusion 2.29 KJ/mol

1 answer:

dmitriy555 [2]3 years ago

4 0

Answer:

$-270.321012\ J$

Explanation:

$C_v$ = Heat capacity of Hg = 28 J/mol

$\Delta T$ = Change in temperature = $(234.32-(273.15+25))$

$\Delta H_{f}$ = Enthalpy of fusion = 2.29 kJ/mol

The number of moles is given by

$n=13.3\times \dfrac{1}{200.59}\\\Rightarrow n=0.0663\ molHg$

Heat is given by

$Q_1=nC_v\Delta T\\\Rightarrow Q_1=0.0663\times 28\times (234.32-(273.15+25))\\\Rightarrow Q_1=-118.494012\ J$

Heat released is given by

$Q_2=-n\Delta H_{f}\\\Rightarrow Q_2=-0.0663\times 2.29\times 10^3\\\Rightarrow Q_2=-151.827\ J$

Total heat is given by

$Q=Q_1+Q_2\\\Rightarrow Q=-118.494012+(-151.827)\\\Rightarrow Q=-270.321012\ J$

The total heat released is $-270.321012\ J$

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<h3>Introduction</h3>

Hi ! Now, I will help to discuss about the gravitational force between two objects. The force of gravity is not affected by the radius of an object, but radius between two object. Moreover, if the object is a planet, the radius of the planet is only to calculate the "gravitational acceleration" on the planet itself,does not determine the gravitational force between the two planets. For the gravitational force between two objects, it can be calculated using the following formula :

$\boxed{\sf{\bold{F = G \times \frac{m_1 \times m_2}{r^2}}}}$

With the following condition :

F = gravitational force (N)
G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_1}$ = mass of the first object (kg)
$\sf{m_2}$ = mass of the second object (kg)
r = distance between two objects (m)

<h3>Problem Solving</h3>

We know that :

G = gravity constant ≈ $\sf{6.67 \times 10^{-11}}$ N.m²/kg²
$\sf{m_X}$ = mass of the planet X = $\sf{1.55 \times 10^{22}}$ kg.
$\sf{m_Y}$ = mass of the planet Y = $\sf{3.95 \times 10^{28}}$ kg.
r = distance between two objects = $\sf{3.75 \times 10^{11}}$ m.

What was asked :

F = gravitational force = ... N

Step by step :

$\sf{F = G \times \frac{m_X \times m_Y}{r^2}}$

$\sf{F = 6.67 \cdot 10^{-11} \times \frac{1.55 \cdot 10^{22} \cdot 3.95 \times 10^{28}}{(3.75 \times 10^{11})^2}}$

$\sf{F \approx \frac{40.84 \times 10^{-11 + 22 + 28}}{14.0625 \times 10^{22}}}$

$\sf{F \approx 2.9 \times 10^{39 - 22}}$

$\sf{F \approx 2.9 \times 10^{17} \: N}$

<h3>Conclusion</h3>

So, the force of gravity that the asteroid and the planet have on each other approximately

$\boxed{\sf{2.9 \times 10^{17} \: N}}$

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