To solve this problem we will apply the concepts given for the efficiency of an engine which is given as
[tex]\eta = 1-\frac{T_C}{T_H}[/tex]
[tex]\eta = \frac{T_H-T_C}{T_H}[/tex]
Where
[tex]T_C[/tex] = Temperature of the cold reservoir
[tex]T_H[/tex] = Temperature of the hot reservoir
The efficiency maximum would be given only if [tex]T_C = 0[/tex]
Replacing this value we have
[tex]\eta = \frac{T_H-0}{T_H}[/tex]
[tex]\eta = 1[/tex]
Therefore: Cold reservoir as cold as possible provide the greater efficiency.
