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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
Physics
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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An archer fires an arrow at a castle 230 m away. The arrow is in flight for 6 seconds, then hits the wall of the castle and stic
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Answer:

(a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

Explanation:

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Horizontal distance = 230 m

Time t = 6 sec

Vertical distance = 16 m

We need to calculate the horizontal component

Using formula of horizontal component

R =u\cos\theta t

Put the value into the formula

\dfrac{230}{6} = u\cos\theta

u\cos\theta=38.33.....(I)

We need to calculate the height

Using vertical component

H=u\sin\theta t-\dfrac{1}{2}gt^2

Put the value in the equation

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u\sin\theta=\dfrac{16+9.8\times18}{6}

u\sin\theta=32.06.....(II)

Dividing equation (II) and (I)

\dfrac{u\sin\theta}{u\cos\theta}=\dfrac{32.06}{38.33}

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\theta=\tan^{-1}0.8364

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(a). We need to calculate the initial speed

Using equation (I)

u\cos\theta\times t=38.33

Put the value into the formula

u =\dfrac{230}{6\times\cos39.90}

u=49.96\ m/s

(b). We have already calculate the angle.

Hence, (a). The initial speed of the arrow is 49.96 m/s.

(b). The angle is 39.90°.

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