The greenhouse effect traps the sun's energy on Earth similar to

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

If the ultimate shear strength of steel is taken to be 3.00 109 Pa, what force is required to punch through a steel plate 1.60 c

photoshop1234 [79]

Answer:

Explanation:

Given

Ultimate shear strength $\tau =3\times 10^9\ Pa$

thickness of steel Plate $t=1.60\ cm$

Cross-sectional $A=1.20\times 10^2\ cm^2$

suppose d is the diameter of cross-section

$\frac{\pi }{4}d^2=1.20\times 10^2$

$d=12.359\ cm$

Shear area of circular cross-section

$A_s=\pi dt$

$A_s=\pi \times 12.359\times 1.6$

$A_s=62.13\ cm^2$

Shear Force $F_s=A_s\times \tau$

$F_s=62.13\times 10^2\times 3000$

$F_s=18.639\ MN$

6 0

3 years ago

A man starts his car from rest and accelerates at 1 m/s2 for 2 s. He then continues at a constant velocity for 10 s until

Bumek [7]

Answer: I showed you all calculation . You did not attach any graph to question .

Explanation:

Lets first find Velocity

Vr=o m/s

Ve=?

a=1m/s²

t=2s

----------

a=(Vr-V)/t

1m/s²=Vr-0m/s/2s

2m/s=Vr

Lets find the time neeeded to stop :

a=1m/s²

Vs=2m/s

Vf=0m/s

a=(Vf-Vs)/t

t*1m/s²=2m/s

t=2 s

4 0

3 years ago

Read 2 more answers

If the graph shown is a position-time graph of an object moving at constant velocity, what is the velocity of the object? Assume

belka [17]

Answer:

C.) v = 50 m/s

Explanation:

The relationship between position vs. time graph and velocity is that the derivative (slope) of the position vs. time graph gives you velocity. In other words, find the slope to get the velocity.

Select two arbitrary points. I'll choose (3s, 225m) and (0s, 75m).

Now use the slope equation:

$v = \frac{y_{2} - y_{1} }{x_{2} - x_{1}} =\frac{225 - 75}{3-0} = 50 m/s$

v = 50 m/s

Hopes this helps!

5 0

3 years ago

How did scientists determine the age of the ocean floor?

mihalych1998 [28]

I believe it's magnetic stripping. I hope that helps

4 0

4 years ago