Two capacitors are connected in parallel through a switch. c1= 1uf, c2=0.25uf. initially the switch is open, c1 is charged to 10v. what happens if we close the switch?

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If the heat engine is a perfect carnot cycle engine, thenefficiency = (t2-t1)/t2   where t1=temperature of cold reservoir ( ° k) = 135            t2=temperature of hot reservoir ( ° k)   = 325thereforeefficiency = (t2-t1)/t2 = (325-135)/325 = 0.5846 = 58%  note: this is the theoretical maximum efficiency of a perfect heat engine.   real-life engines do not achieve this value.