Aheat engine receives an amount of energy qh= 790 kj by heat transfer from a high temperature thermal reservoir at th=950 k. energy is rejected by heat transfer to a lower temperature thermal reservoir at t1=590 k. if waste heat in the amount of q1=160 kj is rejected to the low temperature thermal reservoir during each cycle.

a) solve for the maximum theoretical efficiency that an engine in this situation could operate with. 0.379

b) solve for actual efficiency that the engine is operating with.

c) which of the following best describes the manner in which the cycle is

-reversibly or impossibly?

(a)

(b)

(c) Impossibly

Explanation:

Given:

(a)

Now the maximum theoretical efficiency of the engine:

(b)

Actual efficiency of the heat engine:

(c)

This is impossible because the actual efficiency can never be greater than the ideal (Carnot) efficiency of a heat engine.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.