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kondaur [170]
4 years ago
15

A metal ball of mass 100 g is heated to 90°C and then cooled to 25°C. The heat lost in the process is 2.5 kJ. Another metal ball

of mass 200 g is heated to 90°C and then cooled to 25°C. The heat lost in the process is 5.0 kJ. What can be concluded from the data?
Physics
2 answers:
Masteriza [31]4 years ago
8 0

Answer:

( B )  The first ball is made of a material with a higher specific heat.

Explanation:

kobusy [5.1K]4 years ago
3 0

Explanation:

q = mCΔT

where q is heat,

m is mass,

C is specific heat capacity,

and ΔT is temperature change.

For the first ball:

2500 J = (100 g) C (90°C − 25°C)

C = 0.385 J/g/°C

For the second ball:

5000 J = (200 g) C (90°C − 25°C)

C = 0.385 J/g/°C

The two metals have the same specific heat, and are likely the same metal (possibly copper or zinc).

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A gas mixture contains 320mg methane, 175 mg argon, 225 mg nitrogen (N2). The partial pressure of argon at 300K is 12.52 kPa. Wh
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<u>Answer:</u> The volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For methane:</u>

Given mass of methane = 320 mg = 0.3 g     (Conversion factor:  1 g = 1000 mg)

Molar mass of methane = 16 g/mol

Putting values in equation 1, we get:

\text{Moles of methane}=\frac{0.3g}{16g/mol}=0.019mol

  • <u>For argon:</u>

Given mass of argon = 175 mg = 0.175 g

Molar mass of argon = 40 g/mol

Putting values in equation 1, we get:

\text{Moles of argon}=\frac{0.175g}{40g/mol}=0.0044mol

  • <u>For nitrogen:</u>

Given mass of nitrogen = 225 mg = 0.225 g

Molar mass of nitrogen = 28 g/mol

Putting values in equation 1, we get:

\text{Moles of nitrogen}=\frac{0.225g}{28g/mol}=0.0080mol

To calculate the volume of the mixture, we use the equation:

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We are given:

Partial pressure of argon = 12.52 kPa

Temperature = 300 K

R = Gas constant = 8.31\text{L kPa }mol^{-1}K^{-1}

n = number of moles of argon = 0.0044 moles

Putting values in equation 2, we get:

12.52kPa\times V=0.0044mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\V=\frac{0.0044\times 8.31\times 300}{12.52}=0.876L

Now, calculating the total pressure of the mixture by using equation 2:

Total number of moles = [0.019 + 0.0044 + 0.0080] mol = 0.0314 mol

V= volume of the mixture = 0.876 L

Putting values in equation 2, we get:

P\times 0.876L=0.0314mol\times 8.31\text{L kPa }mol^{-1}K^{-1}\times 300K\\\\P=\frac{0.0341\times 8.31\times 300}{0.876}=89.36kPa

Hence, the volume and total pressure of the mixture is 0.876 L and 89.36 kPa respectively.

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