Respuesta :
Answer:
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:
[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]
were [tex]T_H[/tex] and [tex]T_C[/tex] 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:
[tex]T_C= 19.4+273.15=292.55 K[/tex]
Now, if we know the Temperature of the cold reservoir, and the engine's efficiency, we can use that information to get [tex]T_H[/tex] as follows:
[tex]\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[/tex]
Now, that is the Temperature of the hot reservoir to begin with, but if we want a higher efficency, we need to increase [tex]T_H[/tex] until [tex]\eta=0.623[/tex].
To find out what value [tex]T_H[/tex] has to reach, we repeat the same calculation as before, except now [tex]\eta= 0.623[/tex]
[tex]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[/tex]
So [tex]T_H[/tex] has increased from [tex]416.14K[/tex] to [tex]776 K[/tex], that means an increase in temperature of :[tex]\DeltaT= 776K-416.14K=359.86K[/tex]
This increment of 359.86K is also an increment of 359.86°C, because increments are relative measures of temperature.
Answer:
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
[tex]\eta =1-\frac{T_{2}}{T_{1}}[/tex]
[tex]0.297 =1-\frac{292.4}{T_{1}}[/tex]
[tex]0.703=\frac{292.4}{T_{1}}[/tex]
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
[tex]\eta =1-\frac{T_{2}}{T_{1}}[/tex]
[tex]0.0.623 =1-\frac{292.4}{T'_{1}}[/tex]
[tex]0.377=\frac{292.4}{T'_{1}}[/tex]
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
