Answer:
0.035 (3.5 %)
Explanation:
The thermodynamic efficiency is given by:
[tex]\eta = 1 - \frac{T_C}{T_H}[/tex]
where
[tex]T_C[/tex] is the cold temperature
[tex]T_H[/tex] is the hot temperature
In this problem we have
[tex]T_C = 5 ^{\circ}C+ 273 = 278 KT_H = 15^{\circ}C+273 = 288 K[/tex]
So the efficiency is
[tex]\eta = 1 - \frac{278 K}{288 K}=0.035[/tex]
