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3 years ago

14

In a quasi-static isobaric expansion, the gas does 26.4 J of work. The initial pressure and volume of the gas is 0.9 atm and 19

L. If the internal energy of the gas increases by 80 J in the expansion, how much heat does the gas absorb?

1 answer:

Gelneren [198K]3 years ago

7 0

Answer:

Q = 106.4 J

Explanation:

Given that

Process is isobaric.

Work done by gas (W)= 26. 4 J

Volume of the gas at initial condition = 0.9 L

Volume of the gas at final condition = 19 L

Increase in the internal energy of gas = 80 J

we know that from first law of thermodynamics

Q = ΔU + W

Here Q is the heat absorb by gas

ΔU change in internal energy of gas

W is the work done by gas

So now by putting the values

Q = ΔU + W

Q = 80 + 26.4

Q = 106.4 J

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A small sphere of reference-grade iron with a specific heat of 447 J/kg K and a mass of 0.515 kg is suddenly immersed in a water

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Answer:

The specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

Explanation:

Let suppose that sphere is cooled down at steady state, then we can estimate the rate of heat transfer ( $\dot Q$ ), measured in watts, that is, joules per second, by the following formula:

$\dot Q = m\cdot c\cdot \frac{T_{f}-T_{o}}{\Delta t}$ (1)

Where:

$m$ - Mass of the sphere, measured in kilograms.

$c$ - Specific heat of the material, measured in joules per kilogram-degree Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures of the sphere, measured in degrees Celsius.

$\Delta t$ - Time, measured in seconds.

In addition, we assume that both spheres experiment the same heat transfer rate, then we have the following identity:

$\frac{m_{I}\cdot c_{I}}{\Delta t_{I}} = \frac{m_{X}\cdot c_{X}}{\Delta t_{X}}$ (2)

Where:

$m_{I}$ , $m_{X}$ - Masses of the iron and unknown spheres, measured in kilograms.

$\Delta t_{I}$ , $\Delta t_{X}$ - Times of the iron and unknown spheres, measured in seconds.

$c_{I}$ , $c_{X}$ - Specific heats of the iron and unknown materials, measured in joules per kilogram-degree Celsius.

$c_{X} = \left(\frac{\Delta t_{X}}{\Delta t_{I}}\right)\cdot \left(\frac{m_{I}}{m_{X}} \right) \cdot c_{I}$

If we know that $\Delta t_{I} = 6.35\,s$ , $\Delta t_{X} = 4.59\,s$ , $m_{I} = 0.515\,kg$ , $m_{X} = 1.263\,kg$ and $c_{I} = 447\,\frac{J}{kg\cdot ^{\circ}C}$ , then the specific heat of the unknown material is:

$c_{X} = \left(\frac{4.59\,s}{6.35\,s} \right)\cdot \left(\frac{0.515\,kg}{1.263\,kg} \right)\cdot \left(447\,\frac{J}{kg\cdot ^{\circ}C} \right)$

$c_{X} = 131.750\,\frac{J}{kg\cdot ^{\circ}C}$

Then, the specific heat of the unknown material is 131.750 joules per kilogram-degree Celsius.

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Answer:

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Explanation:

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to find out

determine the current

solution

we know force F = I×L×B

here I is current and L is rail separation and B is magnetic field

so F = I ×2×2 = 4 I

so

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a = $\frac{4I}{1000}$ m/s²

so equation of motion

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solve we get I

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