Answer:
The electricity cost will be Option d ($0.011).
Explanation:
The given values are:
1 ton coal = 28 million Btu/ton
Or,
1 metric tonne = 28 × 10⁶ Btu
As we know,
Coal's cost,
⇒ [tex]C=30/metric \ tonne[/tex]
[tex]=\frac{30}{28\times 10^6}[/tex]
⇒ [tex]H=10300 \ btu/Kwh[/tex]
Now,
Electricity's cost,
⇒ [tex]C\times H[/tex]
On substituting the estimated values, we get
⇒ [tex]\frac{30}{28\times 10^6}\times 10300[/tex]
⇒ [tex]0.011/ Kwh[/tex]
A piston-cylinder device contains 0.1 kg of hydrogen gas (PG model: cv=10.18, k = 1.4, R= 4.12 kJ/kg-K) at 1000 kPa and 300 K. The gas undergoes an expansion process and the final conditions are 500 kPa, 270 K. If 10 kJ of heat is transferred into the gas from the surroundings at 300 K, determine (a) the boundary work (Wb), and (b) the entropy generated (Sgen) during the process
Answer:
(a) 151.84 kJ
(b) 2.922 kJ/K
Explanation:
(a) The parameters given are;
Mass of hydrogen gas, H₂ = 0.1 kg = 100 g
Molar mass of H₂ = 2.016 g/mol
Number of moles of H₂ = 100/2.016 = 49.6 moles
V₁ = mRT/P = 0.1×4.12×300/1000 = 0.1236 m³
P₁/P₂ = (V₂/V₁)^k
V₂ = (P₁/P₂)^(1/k)×V₁ =0.1236 × (1000/500)^(1/1.4) = 0.3262 m³
Boundary work done = (V₂ - V₁)(P₂ + P₁)/2 = (0.3262 - 0.1236)*(500 + 1000)/2 = 151.84 kJ
(b) Entropy generated ΔS = Cv · ㏑(T₂/T₁) + R ·㏑(v₂/v₁)
=10.18 × ㏑(270/300) + 4.12 ·㏑(0.3262/0.1236) = 2.922 kJ/K.
Which material is used in a photovoltaic cell? A photovoltaic cell converts solar radiation into electric current using a thin sheet of____ like silicon.
Answer:
first on is silicon
Explanation:
Answer:
Silicon
Explanation:
Copper indium dieseline (CIS), cadmium telluride (CdTe), and thin-film silicon are certain polycrystalline thin film materials often used, whereas high-efficiency material such as gallium arsenide (GaAs) often comprise single-crystalline thin film materials3.