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chubhunter [2.5K]

3 years ago

11

An aluminum cup contains 225 g of water and a 40 g copper stirrer, all at 27°C. A 410 g sample of silver at an initial temperatu

re of 88°C is placed in the water. The stirrer is used to stir the mixture gently until it reaches its final equilibrium temperature of 32°C. Calculate the mass of the aluminum cup.

1 answer:

fomenos3 years ago

5 0

Answer:

130.22 g

Explanation:

Parameters given:

Mass of water Mw = 225 g

Mass of stirrer Ms = 40 g

Mass of silver M(S) = 410 g

By applying the law of conservation of energy:

(McCc + MsCs + MwCw)ΔTw = M(S)C(S)ΔT(S)

where Mc = Mass of cup

Cc = Specific heat capacity of aluminium cup = 900 J/gC

Cs = Specific heat capacity of copper stirrer = 387 J/gC

Cw = Specific heat capacity of water = 4186 J/gC

ΔTw = change in temperature of water = 32 - 27 = 5 °C

C(S) = Specific heat capacity of silver = 234 J/gC

ΔT(S) = change in temperature of silver = 88 - 32 = 56 °C

Therefore:

[(Mc * 900) + (40 * 387) + (225 * 4186)] * 5 = 410 * 234 * 56

(900Mc + 957330) * 5 = 5276700

900Mc + 957330 = 5276700 / 5 = 1074528

900Mc = 1074528 - 957330

900Mc = 117198

Mc = 117198/ 900

Mc = 130.22 g

The mass of the cup is 130.22 g.

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Maslowich

Answer:

Half

Explanation:

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Now, if the satellite is moved to a distance $R=4R_E$

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<u>after the satellite is moved, we have:</u>

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3 0

4 years ago

PL-1) A spring that hangs vertically is 25 cm long when no weight is attached to its lower end. Steve adds 250 g of mass to the

Anuta_ua [19.1K]

Answer:

20.42 N/m

Explanation:

From hook's law,

F = ke ......................... Equation 1

Where F = Force applied to the spring., k = spring constant, e = extension.

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k = F/e ................. Equation 2

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F = W = mg.

k = mg/e................ Equation 3

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3 years ago

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

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(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

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Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

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Using formula of lens

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Put the value into the formula

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Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

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