Answer:
55.96kJ
Explanation:
Energy = mass of diethyl ether × enthalpy of vaporization of diethyl ether
Volume (v) = 200mL, density (d) = 0.7138g/mL
Mass = d × v = 0.7138 × 200 = 142.76g
Enthalpy of vaporization of diethyl ether = 29kJ/mol
MW of diethyl ether (C2H5)2O = 74g/mol
Enthalpy in kJ/g = 29kJ/mol ÷ 74g/mol = 0.392kJ/g
Energy = 142.76g × 0.392kJ/g = 55.96kJ
Answer:
smaller one
Explanation:
even though he is moving quicker doesn't mean he will be packing more force in the collision
Answer:
5.43 x 10^-3 Nm
Explanation:
N = 52.5, radius, r = 5.35 cm = 0.0535 m, B = 0.455 T, I = 25.3 mA = 0.0253 A
Torque = N I A B Sin theta
Here, theta = 90 degree
Torque = 52.5 x 0.0253 x 3.14 x 0.0535 x 0.0535 x 0.455
Torque = 5.43 x 10^-3 Nm
Answer:
The work and heat transfer for this process is = 270.588 kJ
Explanation:
Take properties of air from an ideal gas table. R = 0.287 kJ/kg-k
The Pressure-Volume relation is <em>PV</em> = <em>C</em>
<em>T = C </em> for isothermal process
Calculating for the work done in isothermal process
<em>W</em> = <em>P</em>₁<em>V</em>₁ ![ln[\frac{P_{1} }{P_{2} }]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7BP_%7B1%7D%20%7D%7BP_%7B2%7D%20%7D%5D)
= <em>mRT</em>₁ [∵<em>pV</em> = <em>mRT</em>]
= (5) (0.287) (272.039) ![ln[\frac{2.0}{1.0}]](https://tex.z-dn.net/?f=ln%5B%5Cfrac%7B2.0%7D%7B1.0%7D%5D)
= 270.588 kJ
Since the process is isothermal, Internal energy change is zero
Δ<em>U</em> = 
From 1st law of thermodynamics
Q = Δ<em>U </em>+ <em>W</em>
= 0 + 270.588
= 270.588 kJ
Answer:Done doing there job, in the winter they have a break
Explanation:
