a machine has an efficiency of 70 percent, how much work does the machine do when 20000 j of work is done on it a. 14000 b. 2060
1 answer:
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Answer:
- the volume of the second tank is 1.77 m³
- the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Explanation:
Given that;
= 1 m³
= 10°C = 283 K
= 350 kPa
= 3 kg
= 35°C = 308 K
= 150 kPa
Now, lets apply the ideal gas equation;
=
R
=
R
/
The gas constant of air R = 0.287 kPa⋅m³/kg⋅K
we substitute
= ( 3 × 0.287 × 308) / 150
= 265.188 / 150
= 1.77 m³
Therefore, the volume of the second tank is 1.77 m³
Also, =
/ R
= (350 × 1)/(0.287 × 283) = 350 / 81.221
= 4.309 kg
Total mass, =
+
= 4.309 + 3 = 7.309 kg
Total volume =
+
= 1 + 1.77 = 2.77 m³
Now, from ideal gas equation;
=
R
/
given that; final temperature = 20°C = 293 K
we substitute
= ( 7.309 × 0.287 × 293) / 2.77
= 614.6211119 / 2.77
= 221.88 kPa ≈ 222 kPa
Therefore, the final equilibrium pressure of air is 221.88 kPa ≈ 222 kPa
Answer:
Explanation:
As we know by energy conservation
All its kinetic energy will convert into thermal energy to raise its temperature
now divide both sides by mass of the object
so change in temperature is given as