22. With TL = 290 k, we find
TL TL 290 K
µ =1 - =Ò! TH = =
TH 1 - µ 1 - 0.40
which yields the (initial) temperature of the high-temperature reservoir: TH = 483 K. If we replace

µ =0.40 in the above calculation with µ =0.50, we obtain a (final) high temperature equal to TH = 580 K.
The difference is

TH - TH = 580 K - 483 K = 97 K .