Krishne wants to measure the mass and volume of a thimble. Which tools should she use?

A closed, rigid tank fitted with a paddle wheel contains 2 kg of air, initially at 300 K. During an interval of 5 minutes, the p

anzhelika [568]

Answer:

The final temperature of the air is $T_2= 605 K$

Explanation:

We can start by doing an energy balance for the closed system

$\Delta KE+\Delta PE+ \Delta U = Q - W$

where

$\Delta KE$ = the change in kinetic energy.

$\Delta PE$ = the change in potential energy.

$\Delta U$ = the total internal energy change in a system.

Q = the heat transferred to the system.

W = the work done by the system.

We know that there are no changes in kinetic or potential energy, so $\Delta KE = 0$ and $\Delta PE=0$

and our energy balance equation is $\Delta U = Q - W$

We also know that the paddle-wheel transfers energy to the air at a rate of 1 kW and the system receives energy by heat transfer at a rate of 0.5 kW, for 5 minutes.

We use this information to calculate the total internal energy change $\Delta U=W+Q$ using the energy balance equation.

We convert the interval of time to seconds $t = 5 \:min = 300\:s$

$\Delta \dot{U}=\dot{W}+ \dot{Q}\\=\Delta U=(W+ Q)\cdot t$

$\Delta U=(1 \:kW+0.5\:kW)\cdot 300\:s\\\Delta U=450 \:kJ$

We can use the change in specific internal energy $\Delta U = m(u_2-u_1)$ to find the final temperature of the air.

We are given that $T_1=300 \:K$ and the air can be describe by ideal gas model, so we can use the ideal gas tables for air to determine the initial specific internal energy $u_1$

$u_1=214.07\:\frac{kJ}{kg}$

Next, we will calculate the final specific internal energy $u_2$

$\Delta U = m(u_2-u_1)\\\frac{\Delta U}{m} =u_2-u_1$

$\frac{\Delta U}{m} =u_2-u_1\\u_2=u_1+\frac{\Delta U}{m}$

$u_2=214.07 \:\frac{kJ}{kg} +\frac{450 \:kJ}{2 \:kg}\\u_2= 439.07 \:\frac{kJ}{kg}$

With the value $u_2=439.07 \:\frac{kJ}{kg}$ and the ideal gas tables for air we make a regression between the values $u = 434.78 \:\frac{kJ}{kg},T=600 \:K$ and $u = 442.42 \:\frac{kJ}{kg}, T=610 \:K$ and we find that the final temperature $T_2$ is 605 K.

3 0

3 years ago

the guage pressure in a car tire is 30.0 psi when the air temperature is 0 C as the day warms up and brighten sun shines What is

Viktor [21]

Answer:

$P_2 = 33.297\ psi$

Explanation:

give,

Gauge pressure of car, P₁ = 30 psi

temperature,T₁ = 0° C = 0 + 273 = 273 K

Assuming temperature at the noon = 30° C

T₂ = 30 + 273 = 303 K

Pressure at this temperature, P₂ = ?

Using ideal gas equation

$\dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}$

taking volume as in compressible V₁ = V₂

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}$

$\dfrac{30}{273}=\dfrac{P_2}{303}$

$P_2 = 303\times \dfrac{30}{273}$

$P_2 = 33.297\ psi$

Hence, Pressure of the at 30°C is equal to 33.297 psi.

3 0

3 years ago

Spinning situations???

Papessa [141]

Just like mass, energy, linear momentum, and electric charge, angular momentum is also conserved.

The wheel has angular momentum. I don't remember whether it's
up or down (right-hand or left-hand rule), but it's consistent with
counterclockwise rotation as viewed from above.

When you grab the wheel and stop it from spinning (relative to you),
that angular momentum has to go somewhere.

As I see it, the angular momentum transfers through you as a temporary
axis of rotation, and eventually to the merry-go-round. Finally, all the mass
of (merry-go-round) + (you) + (wheel) is rotating around the big common
axis, counterclockwise as viewed from above, and with the magnitude
that was originally all concentrated in the wheel.

7 0

3 years ago

Read 2 more answers

If it takes 150N of force to move a box 10 meters. what is the work done on the box?

sergeinik [125]

Answer:

1500 Joules

Explanation:

Work = Force x Distance

When multiplying by 10 you simply shift all the digits to the

left and append a 0 to the end.

so 150 x 10 = 1500 Joules

3 0

3 years ago

Read 2 more answers

A ball, with a mass of 5.9kg, is thrown directly upwards. It reaches a maximum height of 10m from the point at which it was rele

katrin2010 [14]

Answer:

14 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, when the ball is thrown from the ground, it has only kinetic energy, which is given by

$K=\frac{1}{2}mv^2$

where m = 5.9 kg is the mass of the ball and v is its initial speed.

As the ball goes up, its speed decreases, so its kinetic energy decreases and converts into gravitational potential energy. When the ball reaches its maximum height, the speed has become zero, and all the kinetic energy has been converted into gravitational potential energy, given by:

$U=mgh$

where g = 9.8 m/s^2 is the gravitational acceleration and h = 10 m is the maximum height reached by the ball.

Since we can ignore air resistance, energy must be conserved, so the initial kinetic energy must be equal to the final potential energy of the ball, so we can write:

$K=U\\\frac{1}{2}mv^2=mgh$

And we can solve the equation to find v, the initial speed of the ball:

$v=\sqrt{2gh}=\sqrt{2(9.8 m/s^2)(10 m)}=14 m/s$

8 0

3 years ago