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3 years ago

Determine the work done by 2. 00 kg of water when it is all boiled to steam at 100 ∘c. Assume a constant pressure of 1. 00 atm.

Physics

1 answer:

neonofarm [45]3 years ago

4 0

The work done by the water at the constant pressure is $-4.52 \times 10^6 \ J$ .

<h3>Heat required to boil the water</h3>

The heat required to boil the water is calculated as follows;

$Q = mL_v\\\\$

where;

$L_v$ is the heat of vaporization of water = 22.6 x 10⁵ J/kg
m is mass of water

$Q = 2 \times 22.6 \times 10^5\\\\Q = 4.52 \times 10^6 \ J$

<h3>Work done by the water</h3>

The work done by the water at the constant pressure is equal to the heat gained by the water.

$W = - \Delta H\\\\W = -4.52 \times 10^6 \ J$ .

Learn more about first law of thermodynamics here: brainly.com/question/2965070

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Calculate the pressure exerted by 11.1 moles of neon gas in a volume of 5.45 L at 25°C using (a) the ideal gas equation and (b)

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Answer:

49.82414 atm

50.74675 atm

Explanation:

P = Pressure

V = Volume = 5.45 L

R = Gas constant = 0.08205 L atm/mol K

T = Temperature = 25°C

a = 0.211 atm L²/mol²

b = 0.0171 atm L²/mol²

From ideal gas law we have

$PV=nRT\\\Rightarrow P=\dfrac{nRT}{V}\\\Rightarrow P=\dfrac{11.1\times 0.08205(273.15+25)}{5.45}\\\Rightarrow P=49.82414\ atm$

The pressure is 49.82414 atm

From Van der Waals equation we have

$\left(P+\frac{an^2}{V^2}\right)\left(v-nb\right)=nRT\\\Rightarrow P=\dfrac{nRT}{V-nb}-\dfrac{an^2}{V^2}\\\Rightarrow P=\dfrac{11.1\times 0.08205\times (273.15+25)}{5.45-(11.1\times 0.0171)}-\dfrac{0.211\times 11.1^2}{5.45^2}\\\Rightarrow P=50.74675\ atm$

The pressure is 50.74675 atm

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Any child is pushing a shopping cart at a speed of 1.5 m/s.how long will it take this child to push the cart down the aisle with

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Answer:

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3 years ago

Read 2 more answers

An object with a mass of 0. 25 kg is undergoing simple harmonic motion at the end of a vertical spring with a spring constant, k

skelet666 [1.2K]

Answer:

1) The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object at any point of its motion is 16.892 joules.

Explanation:

1) An object under simple harmonic motion is conservative, since there is no dissipative forces acting during motion (i.e. friction, air viscosity). The amplitude of the motion can be found easily by Principle of Energy Conservation by the fact that maximum elastic potential energy ( $U_{e}$ ), in joules, is equal to maximum translational kinetic energy ( $K$ ), in joules:

$U_{e} = K$

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$ (1)

Where:

$k$ - Spring constant, in newtons per meter.

$A$ - Amplitude, in meters.

$m$ - Object mass, in kilograms.

$v$ - Speed of the object at equilibrium, in meters per second.

If we know that $k = 450\,\frac{N}{m}$ , $m = 0.25\,kg$ and $v = 0.3\,\frac{m}{s}$ , then the amplitude of the motion is:

$\frac{1}{2}\cdot k \cdot A^{2} = \frac{1}{2}\cdot m \cdot v^{2}$

$k\cdot A^{2} = m\cdot v^{2}$

$A = v\cdot \sqrt{\frac{m}{k} }$

$A = \left(0.3\,\frac{m}{s} \right)\cdot \sqrt{\frac{0.25\,kg}{0.3\,\frac{m}{s} } }$

$A \approx 0.274\,m$

The amplitude of the motion is approximately 0.274 meters.

2) The total energy of the object ( $E$ ), in joules, is found either by maximum elastic potential energy or by maximum translational kinetic energy, that is: ( $k = 450\,\frac{N}{m}$ , $A \approx 0.274\,m$ )

$E = U_{e}$

$E = \frac{1}{2}\cdot k\cdot A^{2}$

$E = \frac{1}{2}\cdot \left(450\,\frac{N}{m} \right) \cdot (0.274\,m)^{2}$

$E = 16.892\,J$

The total energy of the object at any point of its motion is 16.892 joules.

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