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Naily [24]
2 years ago
7

An iron bar at 200°C is placed in thermal contact with an identical iron bar at 120°C in an at isolated system. After 30 minutes

, both iron bars are at
160°C. If the iron bars were placed in thermal contact in an open system instead of an isolated system, how would the results be different?
Assume that the ropm temperature is 25°C.

A.) The temperatures of the iron bars after 30 minutes would be less than 160°C because heat would be lost to the surroundings.

B.) It would take more than 30 minutes for both iron bars to reach 160°C because heat would be transferred less efficiently.

C.) The temperatures of both iron bars would increase as they absorb heat from the surroundings.

D.) The temperatures of both iron bars would decrease because pieces of them would be lost to the surroundings.
Chemistry
2 answers:
Mila [183]2 years ago
8 0

Answer:

please don't ask for me I don't have no

stich3 [128]2 years ago
5 0

Answer: The correct answers are A and B.

Explanation:

Conduction: In the conduction, the heat is transferred from the hotter body to the colder body until the temperature on both bodies are equal.

In thermal equilibrium, there is no heat transfer as the heat is transferred till the temperature on the bodies are not same.

In the given problem, an iron bar at 200°C is placed in thermal contact with an identical iron bar at 120°C in an isolated system. After 30 minutes, the thermal equilibrium is attained. Then, the temperature on both iron bars are equal.Both iron bars are at 160°C in an isolated system.

But in an open system, the temperatures of the iron bars after 30 minutes would be less than 160°C. There will be heat lost to the surrounding. The room temperature is 25°C. There will be exchange of the heat occur between the iron bars and the surrounding. But It would take more than 30 minutes for both iron bars to reach 160°C because heat would be transferred less efficiently.

Therefore, the correct options are (A) and (B).

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A gas with a volume of 250 mL at a temperature of 293K is heated to 324K. What is the new volume of the gas?
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Answer:

The new volume of the gas is 276.45 mL.

Explanation:

Charles's law indicates that for a given sum of gas at constant pressure, as the temperature increases, the volume of the gas increases, and as the temperature decreases, the volume of the gas decreases.

Charles's law is a law that mathematically says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

\frac{V}{T} =k

Analyzing an initial state 1 and a final state 2, it is satisfied:

\frac{V1}{T1} =\frac{V2}{T2}

In this case:

  • V1= 250 mL
  • T1= 293 K
  • V2= ?
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Replacing:

\frac{250 mL}{293 K} =\frac{V2}{324 K}

Solving:

V2=324 K*\frac{250 mL}{293 K}

V2= 276.45 mL

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4 0
3 years ago
A large balloon contains 5400 m3 of He gas that is kept at a temperature of 280 K and an absolute pressure of 1.10 x 105 Pa. Fin
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Answer:

1.02 × 10⁶ g

Explanation:

Step 1: Given data

  • Volume of the balloon (V): 5400 m³
  • Temperature (T): 280 K
  • Absolute pressure (P): 1.10 × 10⁵ Pa
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Step 2: Convert "V" to L

We will use the conversion factor 1 m³ = 1000 L.

5400 m³ × 1000 L/1 m³ = 5.400 × 10⁶ L

Step 3: Convert "P" to atm

We will use the conversion factor 1 atm = 101325 Pa.

1.10 × 10⁵ Pa × 1 atm / 101325 Pa = 1.09 atm

Step 4: Calculate the moles of He (n)

We will use the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.09 atm × 5.400 × 10⁶ L / 0.08206 atm.L/mol.K × 280 K

n = 2.56 × 10⁵ mol

Step 5: Calculate the mass of He (m)

We will use the following expression.

m = n × M

m = 2.56 × 10⁵ mol × 4.002 g/mol

m = 1.02 × 10⁶ g

8 0
3 years ago
A marble rolls off horizontally from the edge of table top 1.50 m above the floor. it strikes the floor 2.0 m from the base of t
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Answer:

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