Physics, 05.09.2019 18:30 Brock2002

Acarnot engine whose low-temperature reservoir is at 19.4°c has an efficiency of 29.7%. by how much should the celsius temperature of the high-temperature reservoir be increased to increase the efficiency to 62.3%?

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Answer from: saltytaetae

The Temperature of the Hot reservoir should be increased by 359.86°C

Explanation:

A Carnot engine is a reversible engine, that means that it can work both as a heat engine and as a refrigerator, it can both use heat from a hot reservoir (part of which later must end up in a cold reservoir) to produce work, or use work to move heat from a cod to a hot reservoir.

The efficiency of a carnot engine working as a heat engine is given by:

$\eta = 1 - \frac{T_C}{T_H}$

were T_H and T_C are the Absolute temperatures of the hot and cold reservoir respectively.

To get the absolute temperature in K from the relative temperature in °C, we must add 273.15 K (the absolute temperature of the freezing point of water at atmospheric pressure)

Thus:

T_C= 19.4+273.15=292.55 K

Now, if we know the Temperature of the cold reservoir, and the engine's efficiency, we can use that information to get T_H as follows:

$\eta = 1 - \frac{T_C}{T_H}\\\eta=0.297\\0.297=1 - \frac{292.55K}{T_H}\\\frac{292.55K}{T_H}=1-0.297=0.703\\T_H=\frac{292.55K}{0.703}=416.14K$

Now, that is the Temperature of the hot reservoir to begin with, but if we want a higher efficency, we need to increase T_H until $\eta=0.623$ .

To find out what value T_H has to reach, we repeat the same calculation as before, except now $\eta= 0.623$

$0.623=1 - \frac{292.55K}{T_H}\\\frac{292.55K}{T_H}=1-0.623=0.377\\T_H=\frac{292.55K}{0.377}=776.00K$

So T_H has increased from 416.14K to 776 K , that means an increase in temperature of : $\DeltaT= 776K-416.14K=359.86K$

This increment of 359.86K is also an increment of 359.86°C, because increments are relative measures of temperature.

Answer from: Jasten

359.67°C or 359 K

Explanation:

Case I:

T2 = 19.4°C = 19.4 + 273 = 292.4 K

η = 29.7 % = 0.297

Let T1 be the temperature of hot reservoir.

By using the formula for the efficiency of Carnot's heat engine

$\eta =1-\frac{T_{2}}{T_{1}}$

$0.297 =1-\frac{292.4}{T_{1}}$

$0.703=\frac{292.4}{T_{1}}$

T1 = 415.93 k

Case II:

T2 = 19.4°C = 19.4 + 273 = 292.4 K

η = 62.3 % = 0.623

Let T'1 be the temperature of hot reservoir.

By using the formula for the efficiency of Carnot's heat engine

$\eta =1-\frac{T_{2}}{T_{1}}$

$0.0.623 =1-\frac{292.4}{T'_{1}}$

$0.377=\frac{292.4}{T'_{1}}$

T'1 = 775.6 k

Incraese in the temperature of hot reservoir

= T'1 - T1 = 775.6 - 415.93 = 359.67°C or 359 k

Answer from: Quest

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

Answer from: Quest

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down

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Acarnot engine whose low-temperature reservoir is at 19.4°c has an efficiency of 29.7%. by how much...