3. an engine’s fuel is heated to 2,000 k and the surrounding air is 300 k. calculate the ideal efficiency of the engine. hint: the efficiency (e) of a carnot engine is defined as the ratio of the work (w) done by the engine to the input heat qh : e=w/qh. w=qh – qc, where qc is the output heat. that is, e=1-qc/qh =1-tc/th, where tc for a temperature of the cold reservoir and th for a temperature of the hot reservoir. the unit of temperature must be in kelvin.

E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

Explanation:

The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH

E = W/QH.

W=QH – QC,

Where Qc is the output heat.

That is,

E=1 - Qc/QH

E =1 - Tc/TH

where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.

Note: The unit of temperature must be in Kelvin.

Tc = 300K

TH = 2000K

Substituting the values of E, we have;

E = 1 - 300K/2000K

E = 1 - 0.15

E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

if the person is shown to have +90n, then ans is -10n.

if the person is shown to have -90n, then ans is +10n.

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form