SOMEONE PLEASE HELP MEEEEEEE

how much water is needed to produce 1kwh of electricity at a power plant that is 30% efficient if the temperature increase 10 C

Dimas [21]

The amount of water needed is 287 kg

Explanation:

The amount of energy that we need to produce with the power plant is

$E=1 kWh = (1000W)(1h)=(1000W)(3600s)=3.6\cdot 10^6 J$

We also know that the power plant is only 30% efficient, so the energy produced in input must be:

$E_{in}=\frac{E}{0.30}=\frac{3.6\cdot 10^6}{0.3}=1.2\cdot 10^7 J$

The amount of water that is needed to produce this energy can be found using the equation

$E_{in}=mC\Delta T$

where:

m is the amount of water

$C=4186 J/kg^{\circ}C$ is the specific heat capacity of water

$\Delta T=10^{\circ}C$ is the increase in temperature

And solving for m, we find:

$m=\frac{E_{in}}{C\Delta T}=\frac{1.2\cdot 10^7}{(4186)(10)}=287 kg$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

3 0

3 years ago

A copper wire and a tungsten wire of the same length have the same resistance. What is the ratio of the diameter of the copper w

spayn [35]

Answer:

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

Explanation:

Resistance: Resistance is defined to the ratio of voltage to the electricity.

The resistance of a wire is

directly proportional to its length i.e $R\propto l$
inversely proportional to its cross section area i.e $R\propto \frac{1}{A}$

Therefore

$R=\rho\frac{l}{A}$

ρ is the resistivity.

The unit of resistance is ohm (Ω).

The resistivity of copper(ρ₁) is 1.68×10⁻⁸ ohm-m

The resistivity of tungsten(ρ₂) is 5.6×10⁻⁸ ohm-m

For copper:

$A=\pi r_1^2 =\pi (\frac{d_1}{2} )^2$

$R_1=\rho_1\frac{l_1}{\pi(\frac{d_1}{2})^2 }$

$\Rightarrow (\frac{d_1}{2})^2=\rho_1\frac{l_1}{\pi R_1 }$ ......(1)

Again for tungsten:

$R_2=\rho_2\frac{l_2}{\pi(\frac{d_2}{2})^2 }$

$\Rightarrow (\frac{d_2}{2})^2=\rho_2\frac{l_2}{\pi R_2 }$ ........(2)

Given that $R_1=R_2$ and $l_1=l_2$

Dividing the equation (1) and (2)

$\Rightarrow\frac{ (\frac{d_1}{2})^2}{ (\frac{d_2}{2})^2}=\frac{\rho_1\frac{l_1}{\pi R_1 }}{\rho_2\frac{l_2}{\pi R_2 }}$

$\Rightarrow( \frac{d_1}{d_2} )^2=\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}$ [since $R_1=R_2$ and $l_1=l_2$ ]

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{1.68\times 10^{-8}}{5.6\times 10^{-8}}}$

$\Rightarrow( \frac{d_1}{d_2} )=\sqrt{\frac{3}{10}}$

$\Rightarrow d_1:d_2=\sqrt{3} :\sqrt{10}$

Therefore the ratio of diameter of the copper to that of the tungsten is

$\sqrt{3} :\sqrt{10}$

8 0

3 years ago

Coders play an important role in

Burka [1]

Answer:

it is a.health record documentation

Explanation:hope this helps

3 0

3 years ago

3. In which activity is no work done?

Firlakuza [10]

Answer:

Sleeping, since your brain is working, but your physical body is not moving at all or moving rarely, therefore there is no work done in the activity of sleeping.

Explanation:

6 0

4 years ago

The winch takes in cable at the constant rate of 130 mm/s. if the cylinder mass is 115 kg, determine the tension in cable 1. neg

nikitadnepr [17]

By applying Newton's second law of motion;

ma = mg - T

Where,
m = mass; a = downward accelerations (+ve value) or upward acceleration (-ve value); g = gravitational acceleration; T = tension.

For the current case, the velocity is constant therefore,
a = 0

Then,
0 = mg - T
T = mg = 115*9.81 = 1128.15 N

Tension in the cable is 1128.15 N.

8 0

3 years ago