Answer:
<em>Heat rejected to cold body = 3.81 kJ</em>
Explanation:
Temperature of hot thermal reservoir Th = 1600 K
Temperature of cold thermal reservoir Tc = 400 K
<em>efficiency of the Carnot's engine = 1 - </em>
<em> </em>
eff. of the Carnot's engine = 1 - 
eff = 1 - 0.25 = 0.75
<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>
work done by heat engine = 2 kJ
<em>eff. of heat engine is gotten as = W/Q</em>
where W = work done by heat engine
Q = heat rejected by heat engine to lower temperature reservoir
from the equation, we can derive that
heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>
Answer:
water sample have more water content
Explanation:
given data
soil 1 is saturated with water
unit weight of water = 1 g/cm³
soil 2 is saturated with alcohol
unit weight of alcohol = 0.8 g/cm³
solution
we get here water content that is express as
water content = 
....................1
here soil is full saturated so 
is 100% in both case
so put here value for water
water content = 100 % × 1
water content = 1 g
and
now we get for alcohol that is
water content = 100 % × 0.8
water content = 0.8 g
so here water sample have more water content
Answer:
Explanation:
Given
scale i.e. 
Using Reynolds number similarity
Properties of air
Properties of sea water
Answer:
Relative humidity 48%.
Dew point 74°F
humidity ratio 118 g of moisture/pound of dry air
enthalpy 41,8 BTU per pound of dry air
Explanation:
You can get this information from a Psychrometric chart for water, like the one attached.
You enter the chart with dry-bulb and wet-bulb temperatures (red point in the attachment) and following the relative humidity curves you get approximately 48%.
To get the dew point you need to follow the horizontal lines to the left scale (marked with blue): 74°F
for the humidity ratio you need to follow the horizontal lines but to the rigth scale (marked with green): 118 g of moisture/pound of dry air
For enthalpy follow the diagonal lines to the far left scale (marked with yellow): 41,8 BTU per pound of dry air
