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3 years ago

12

In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

ture reservoir, and does 400 J of work. What is its efficiency?

1 answer:

Talja [164]3 years ago

4 0

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

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Colt1911 [192]

Answer:

There is a loss of fluid in the container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

$\Delta V = \beta V_0 \Delta T$

Where,

$\Delta V =$ Change in volume

$V_0 =$ Initial Volume

$\Delta T =$ Change in temperature

$\beta =$ coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

$\Delta V_T = \Delta V_l - \Delta V_c$

Where,

$\Delta V_l$ = Change in the volume of liquid

$\Delta V_c$ = Change in the volume of copper

Then replacing with the previous equation we have:

$\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T$

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

$\beta_c = 51*10^{-6}/\°C$

$\beta_l = 400*10^{-6}/\°C$

$V_0 = 16L$

$\Delta T = (95\°C-10\°C)$

Replacing we have that,

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

$\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)$

$\Delta V = 0.475L$

Therefore there is a loss of fluid in the container of 0.475L

6 0

3 years ago

Cientists have changed the model of the atom as they have gathered new evidence. One of the atomic models is shown below.

scZoUnD [109]

Answer:

A few of the positive particles aimed at a gold foil seemed to bounce back.

Explanation:

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2 years ago

A 1000 kg car moving at 108 km/h jams on its brakes and comes to a stop. How much work was done by friction?

Nostrana [21]

Answer:

The work done by friction was $-4.5\times10^{5}\ J$

Explanation:

Given that,

Mass of car = 1000 kg

Initial speed of car =108 km/h =30 m/s

When the car is stop by brakes.

Then, final speed of car will be zero.

We need to calculate the work done by friction

Using formula of work done

$W=\Delta KE$

$W=K.E_{f}-K.E_{i}$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{f}^2$

Put the value of m and v

$W=0-\dfrac{1}{2}\times1000\times(30)^2$

$W=-450000
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$W=-4.5\times10^{5}\ J$

Hence, The work done by friction was $-4.5\times10^{5}\ J$

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3 years ago

A sound wave travels through water. What best describes the direction of the water particles?

balu736 [363]

Answer:

In water, the particles are much closer together, and they can quickly transmit vibration energy from one particle to the next.

A water wave is an example of a transverse wave. As water particles move up and down, the water wave itself appears to move to the right or left.

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It is also used to form an amalgam which is the result of its combination with silver or gold. Mercury has been used to mine gold and silver. This application has also been phased out.
Today's use of mercury includes mercury-vapor lamps which are the bright lamps used in high-ways.

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2 years ago

Read 2 more answers