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3 years ago

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A 50 mL graduated cylinder contains 25.0 mL of water. A 42.5040 g piece of gold is placed in the graduated cylinder and the wate

r level rises to 27.2 mL. What is the density of the piece of gold?

1 answer:

Elenna [48]3 years ago

5 0

Answer:

19320 kg/m³

Explanation:

density: This can be defined as the ratio of the mass of a body to its volume. The S.I unit of Density is kg/m³.

The formula of density is given as,

D = m/v ......................... Equation 1.

Where D = Density of the gold, m = mass of the gold, v = volume of the gold.

Note: From Archimedes's Principle, the piece of gold displace an amount of water that is equal to it's volume.

Amount of water displace = 27.2 - 25 = 2.2 mL.

Given: m = 42.504 g = 0.042504 kg, v = 2.2 mL = (2.2/10⁶) m³ = 0.0000022 m³

Substitute into equation 1

D = 0.042504/0.0000022

D = 19320 kg/m³

Hence the density of the piece of gold = 19320 kg/m³

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Answer:

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Explanation:

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Solution:

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DiKsa [7]

Answer:0.061

Explanation:

Given

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$dW=Q(1-\frac{T_C}{T})$

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integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

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$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

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$=\frac{-40+37.548}{-40}$

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steposvetlana [31]

Answer:

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$W=\int\limits^{r_f}_{r_i}{ma\cdot dr} \,\\$

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$W=m\int\limits^{v_f}_{v_i} v \cdot dv \,$

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Answer:

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