# Chapter 19 - Exercises and Problems - Page 361: 12

99.95%

#### Work Step by Step

As we know that $e_{max}=(1-\frac{T_c}{T_h})\times 100\%$ We plug in the known values to obtain: $e_{max}=(1-\frac{2.7}{5800})\times 100\%=99.95\%$

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