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2 years ago

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A heat engine has a maximum possible efficiency of 0.780. If it operates between a deep lake with a constant temperature of-24.8

°C, what is the temperature of the hot reservoir (in Kelvins)?

1 answer:

Volgvan2 years ago

8 0

Answer : The temperature of the hot reservoir (in Kelvins) is 1128.18 K

Explanation :

Efficiency of carnot heat engine : It is the ratio of work done by the system to the system to the amount of heat transferred to the system at the higher temperature.

Formula used for efficiency of the heat engine.

$\eta =1-\frac{T_c}{T_h}$

where,

$\eta$ = efficiency = 0.780

$T_h$ = Temperature of hot reservoir = ?

$T_c$ = Temperature of cold reservoir = $-24.8^oC=273+(-24.8)=248.2K$

Now put all the given values in the above expression, we get:

$\eta =1-\frac{T_c}{T_h}$

$0.780=1-\frac{248.2K}{T_h}$

$T_h=1128.18K$

Therefore, the temperature of the hot reservoir (in Kelvins) is 1128.18 K

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Answer:

The $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

Explanation:

Given :

Current $I = 1 \times 10^{5}$ A

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Time $t = 1$ sec

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$I = \frac{ne}{t}$

Where $n =$ number of charges means in our case number of electrons

$n = \frac{It}{e}$

$n = \frac{1 \times 10^{5} }{1.6 \times 10^{-19} }$

$n = 6.25 \times 10^{23}$

Therefore, $6.25 \times 10^{23}$ electrons are moving through the superconductor per second.

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2 years ago

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The average velocity of the car for the whole journey is 69.57 km/h.

The given parameters:

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The time spent by the before refueling is calculated as follows;

$t = \frac{d}{v} \\\\t_1 = \frac{240}{75} \\\\t_1 = 3.2 \ hours$

The time spent by the car for the remaining journey;

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The total time of the journey is calculated as follows;

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The average velocity of the car for the whole journey is calculated as follows;

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Learn more about average velocity here: brainly.com/question/6504879

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2 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

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Answer:

241.7 s

Explanation:

We are given that

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Initial time= $t_1=6.36 s$

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We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

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Using the formula

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$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

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