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%?

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:

were and 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:

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

Now, that is the Temperature of the hot reservoir to begin with, but if we want a higher efficency, we need to increase until .

To find out what value has to reach, we repeat the same calculation as before, except now

So has increased from   to , that means an increase in temperature of :

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

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

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

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

the right answer is 0.3 meters per second