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sweet [91]
3 years ago
9

A copper block ( 90°C) is placed in contact with a lead block ( 20°C), which is already in contact with an iron block (70°C). Wh

ich of the following will happen?
A. Heat will flow from the copper to the lead and from the iron to the lead until the temperatures are equal .

B. Heat will flow from the iron to the lead to the copper until the temperatures are equal.

C. Heat will flow from the copper to the lead to the iron until the temperatures are equal.

D. Heat will flow from the lead to the copper and to the iron until the temperatures are equal.
Chemistry
2 answers:
Ostrovityanka [42]3 years ago
3 0
<span>B. Heat will flow from the iron to the lead to the copper until the temperatures are equal. because i don't really speak Italian </span><span>
</span>
Radda [10]3 years ago
3 0

Answer: Option (A) is the correct answer.

Explanation:

According to zeroth law of thermodynamics, when two systems are in thermal equilibrium with the third system then all of them are in thermal equilibrium with each other.

Hence, when copper block is placed adjacent to lead block and lead block is placed adjacent to iron block then heat will flow from hotter block to the colder block.

Thus, we can conclude that heat will flow from the copper to the lead and from the iron to the lead until the temperatures are equal.

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marta [7]

Answer:

1) The new pressure of the gas is 500 kilopascals.

2) The final volume is 1.44 liters.

3) Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

Explanation:

1) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that P_{1} = 100\,kPa, V_{1} = 500\,mL and V_{2} = 1000\,mL, then the new pressure of the gas is:

P_{2} = P_{1}\cdot \left(\frac{V_{1}}{V_{2}} \right)

P_{2} = 500\,kPa

The new pressure of the gas is 500 kilopascals.

2) Let suppose that gas experiments an isothermal process. From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that V_{1} = 3.60\,L, P_{1} = 10\,kPa and P_{2} = 25\,kPa then the new volume of the gas is:

V_{2} = V_{1}\cdot \left(\frac{P_{1}}{P_{2}} \right)

V_{2} = 1.44\,L

The final volume is 1.44 liters.

3) From the Equation of State for Ideal Gases we construct the following relationship:

\frac{P_{2}}{P_{1}} = \frac{V_{1}}{V_{2}} (1)

Where:

P_{1}, P_{2} - Initial and final pressure, measured in kPa.

V_{1}, V_{2} - Initial and final pressure, measured in mililiters.

If we know that \frac{P_{2}}{P_{1}} = 3, then the volume ratio is:

\frac{V_{1}}{V_{2}} = 3

\frac{V_{2}}{V_{1}} = \frac{1}{3}

Volume will decrease by approximately 67 %.

4) The Boyle's Laws deals with pressures and volumes.

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There is a gas pressure in a container that is 123kPa what is the pressure in atmospheres ?
Vera_Pavlovna [14]

Answer:

In strict SI units (highly recommended), express n in moles, R is the universal gas constant R=8.314Jmol−K , T is the temperature in Kelvins, and the volume V is in m3 . The resulting pressure P will be in Pa. R=0.082054L−atmmol−K , in which case the pressure is calculated in atm.

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This family (ethane, propane, butane, etc) of materials is likely to have following set of properties.

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The ethane, propane, butane, belong to alkanes family.The alkanes are also considers as saturated hudrocarbons. Ethane is found in gaseous stae Ethane is the second alkane followed by propane followed by butane.

learn about butane

brainly.com/question/14818671

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