Acarnot engine has a lower operating temperature of 14.6 celcius and efficiency of 21.2 %. by how many kelvins should the high operating temperature be increased to achieve an efficiency of 50.9 %?

it should be increased by 220.73K or -52.22Celsius

Explanation:

A Carnot engine has a lower operating temperature of 14.6 Celsius and efficiency of 21.2 %. By how many kelvins should the high operating temperature be increased to achieve an efficiency of 50.9 %?

effi=

Tc=temperature of the cold reservoir in kelvin(14.6+273=287.6

Th=temperature of the hot reservoir

eff= efficiency of a carnot engine(which alway <100%)

effi=

21.2 %=1-Tc/Th

0.212=1-287.6/Th

-.788=-287.6/Th

Th=364.97K

to achieve an efficiency of 50.9%

0.509=1-Tc/Th

-0.491=-287.6/Th

Th=585.74K

taking the difference between the two high temperatures

585.74K-364.97K

220.73K

it should be increased by 220.73K

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer