The efficiency of the device is 30 %
Explanation:
The efficiency of a heat engine is given by:
where
W is the work done by the engine
is the heat in input to the engine
For the device in this problem, we have:
W = 120 J is the work done
is the heat in input
Substituting, we find the efficiency:
which corresponds to an efficiency of 30%.
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Answer:
Density = 1.1839 kg/m³
Mass = 227.3088 kg
Specific Gravity = 0.00118746 kg/m³
Explanation:
Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³
Now, from tables, density of air at 25°C is 1.1839 kg/m³
Now formula for density is;
ρ = mass(m)/volume(v)
Plugging in the relevant values to give;
1.1839 = m/192
m = 227.3088 kg
Formula for specific gravity of air is;
S.G_air = density of air/density of water
From tables, density of water at 25°C is 997 kg/m³
S.G_air = 1.1839/997 = 0.00118746 kg/m³
Answer:
The amount of energy added to rise the temperature Q = 17413.76 KJ
Explanation:
Mass of water = 52 kg
Initial temperature 
= 68 °F = 20° c
Final temperature 
= 212 °F = 100° c
Specific heat of water 
Now heat transfer Q = m × C × ( - )
⇒ Q = 52 × 4.186 × ( 100 - 20 )
⇒ Q = 17413.76 KJ
This is the amount of energy added to rise the temperature.
Answer:
Newton's law of universal gravitation states that every particle attracts every other particle in the universe with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Explanation:
