Q1
ΔQ = mcΔT + mL = 2•4186•10 + 2.26•10⁶•2 = 4.6•10⁶ J
initial volume V=m/ρ(water) = 2/1000 =2•10⁻³ m³
final volume = 3.3 m³
Change in volume ΔV=V(fin) – V(init) =3.3-2•10⁻³ ≈ 3.3 m³.
Work of expansion is W = p ΔV=1.013•10⁵•3.3 = 3.34•10⁵ J
The 2 Law of thermodynamics
ΔQ=ΔU+W
ΔU= ΔQ – W = 4.6•10⁶ - 3.34•10⁵ = 4.266•10⁶ ≈ 4.27 MJ.
Q2
initial volume V=m/ρ(water) = 0.001/1000 =1•10⁻⁶m³
final volume = 1672 cm³ =1672•10⁻⁶m³
Change in volume ΔV ≈ 1672•10⁻⁶m³.
W= p ΔV =1.013•10 ⁵•1672•10⁻⁶=169.3 J
(Q1)calculate the electric power which must be supplied to the filament of a light bulb operating at 3000k. the total surface area of the filament is 0.000008m2 and iits emissivity is 0.92.(a)20.3w (b)33.8w(c)46.4w(d)56.7w (Q2) calculate the change in internal energy of 2kg of water at 90c.when it is changed to 3.30m3of steam at 100c.The whole process occurs at atmospheric pressure. the latent heat of vaporization of water is 226000000j/kg(a)4.27mj(b)3.43kj(c)45.72mj(d)543.63j (Q3)calculate the work done against external atmospheric pressure when 1g of water changes to 1672Cm3 of steam. Take the atmospheric pressure as 1.01300000 Nm-2 (a)169.3j(b)342.4j(c)226.2j(d)143.5j
1 answer
1 answer