Question

1s the temperature of 3. In Figure 1, the efficiency of a Carnot engine is listed as a function of the hot-reservoir temperature. The cold-temperature reservoir is unknown TH (K) 300 370 425 483 535 0.133 0.298 0.390 0.461 0.513 Figure 1: This is the figure for problem 3

Answer 1

For Carnot Engine, efficiency is given by $\dpi{80} \bg_white \fn_jvn e=(1-\frac{T_{C}}{T_{H}})$

To fit the given data, the data is entered in Excel and line plot is done over it to get the following graph

Carnot Efficiency as function of TH 0.6 0.5 0.4 3 0.3 0.2 0.1 300 370 425 483 535 TH (in Kelvin)

The temperature in cold reservoir can be found out by putting values of T_H and e in the equation

$\dpi{80} \bg_white \fn_jvn 0.133=(1-\frac{T_{C}}{300})$

$\dpi{80} \bg_white \fn_jvn \frac{T_{C}}{300}=(1-0.133)=0.867$

$\dpi{80} \fn_jvn T_{C}=0.867*300=260.1$

Similarly check for other data

$\dpi{80} \fn_jvn T_{C}=(1-0.298)*370=259.74$

$\dpi{80} \fn_jvn T_{C}=(1-0.390)*425=259.25$

$\dpi{80} \fn_jvn T_{C}=(1-0.461)*483=260.33$

$\dpi{80} \fn_jvn T_{C}=(1-0.513)*535=260.545$

All of the above values are approximately same and hence the temperature of cold reservoir can be taken safely as close to 260 Kelvin.