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2 years ago

What does the term "heat capacity" refer to?

2 answers:

5 0

The correct answer is c) part i.e.The amount of heat energy needed to raise the temperature of a unit mass of a material one degree

kenny6666 [7]2 years ago

4 0

Answer:

option C is correct

.............,

You might be interested in

If the bond enthalpy for a C-H bond is 413 kJ, what will happen when the C-H bond is broken?

pentagon [3]

Explanation:

Bond Enthalpy : It is defined as amount of energy required to break a the particular bond in there gaseous state. It is also known as bond energy. It units are kJ/mol.

Breaking of a bond is an Endothermic process (energy absorbed from the surroundings).

Formation of bond is an Exothermic process (energy is released to the surroundings).

If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.

8 0

3 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

The windshield of a car has a total length of arm and blade of 9 inches, and rotates back and forth through an angle of 93degre

Sergio039 [100]

Answer:

The area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

Explanation:

Given

Length of blade = 9 inches

Angle of rotation = 93°

We're to calculate the area of the portion of the windshield cleaned by the 7-in wiper blade?

We'll solve this by using area of a sector.

Area of a sector = ½r²θ

where θ is in radians.

So, angle of rotation (93°) must first be converted to radians

Converting 93º to radians, we get 31π/60

The area of the region swept out by the wiper blade = (area of the sector where r = 9 and

θ = 31π/60) - (area of the sector where r = (9-7) and θ = 31π/60).

We're making use of 9-7 because that region is outside the boundary of the 7in blade

So Area = ½*9²*31π/60 - ½*2²*31π/60

Area = ½*31π/60(9²-7²)

Area = 31π/120 * (81 - 49)

Area = 31π/120 * 32

Area = 992π/120

Area = 62.49151386765697 in²

Area ≈ 62.49 in²

Hence, the area of the portion of the windshield cleaned by the 7-in wiper blade is 62.49 in²

5 0

3 years ago

Read 2 more answers

Which of the following is the closest to the scientific fact

Finger [1]

B - A theory seems to be the closest

7 0

2 years ago

Please help if you can

denis23 [38]

Answer:

The answer is B

Explanation:

Because when the both sides aren't balanced one side has to cause motion. (fall down)

6 0

2 years ago