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vitfil [10]
2 years ago
5

A chemist heats the block of copper as shown in the interactive, then places the metal sample in a cup of oil at 25.00 °C instea

d of a cup of water. The temperature of the oil increases to 27.33 °C . Calculate the mass of oil in the cup. The specific heat of copper is 0.387 J/g⋅°C and the specific heat of oil is 1.74 J/g⋅°C .
Chemistry
1 answer:
Mazyrski [523]2 years ago
6 0

When the oil is added to the heated copper, the energy in the system is

conserved.

  • The mass of the oil in the cup, is approximately <u>64.73 grams</u>.

Reasons:

The question parameters are;

Temperature of the oil in the cup = 25.00°C

Final temperature of the oil and copper, T₂ = 27.33 °C

Specific heat of copper, c₂ = 0.387 J/(g·°C)

Specific heat capacity of oil, c₁ = 1.74 J/(g·°C)

Required:

The<em> mass of oil</em> in the cup.

Solution:

The mass of the copper, m₂ = 17.920 g

Temperature of copper after heating, T₂ = 65.17°C

Temperature of the copper after being placed in the cup of oil, T₂ = 27.33°C

Heat lost by copper = Heat gained by the oil

  • m₂·c₂·(T₂ - T₃) = m₁·c₁·(T₃ - T₁)

Therefore, we get;

17.920 × 0.387 × (65.17 - 27.33) = m₁ × 1.74 × (27.33 - 25)

262.4219136 = 4.0542·m₁

m₁ ≈ 64.73

  • The mass of the oil in the cup, m₁ ≈ <u>64.73 g</u>

Learn more here:

brainly.com/question/21406849

<em>Possible part of the question obtained from a similar question online, are;</em>

<em>The mass of the copper, m₂ = 17.920 g</em>

<em>Temperature of copper after heating = 65.17°C</em>

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Answer:

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Explanation:

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The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

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Now we know three variables:

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Substitute these values into the formula.

4500.0 \ J = m (0.4494 \ J/g \textdegree C)(30.1 \textdegree C)

Multiply on the right side of the equation. The units of degrees Celsius cancel.

4500.0 \ J = m (13.52694 J/g)

We are solving for the mass, so we must isolate the variable m. It is being multiplied by 13.52694 Joules per gram. The inverse operation of multiplication is division, so we divide both sides by 13.52694 J/g

\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

332.6694729 \ g =m

The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

333 \ g \approx m

The mass of the sample of metal is approximately <u>333 grams.</u>

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bagirrra123 [75]
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3 years ago
Strontium-90 is a radioisotope that will decrease in mass by one-half every 29 years. How many years will it take for a 10.0-gra
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KengaRu [80]

Answer:

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Explanation:

Hello!

In this case, since the definition of entropy in a random mixture is:

\Delta S=-n_TR\Sigma[x_i*ln(x_i)]

For this silver-gold mixture we write:

\Delta S=-(n_{Au}+n_{Ag})R\Sigma[\frac{n_{Au}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Au}}{n_{Au}+n_{Ag}} )+\frac{n_{Ag}}{n_{Au}+n_{Ag}} *ln(\frac{n_{Ag}}{n_{Au}+n_{Ag}} )]

By knowing the moles of gold:

n_{Au}=100g*\frac{1mol}{197g} =0.508mol

It is possible to write the aforementioned formula in terms of the variable x representing the moles of silver:

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Which can be solved via Newton-Raphson or a solver software, in this case, I will provide you the answer:

x=n_{Ag}=21.0molAg

So the mass is:

m_{Ag}=21.0mol*\frac{107.9g}{1mol}\\ \\m_{Ag}=2,265.9g

Best regards!

3 0
2 years ago
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