THERMODYNAMICS

Answer :

(C/5) = [(F – 32) / 9]

                   C = 25°C hence

                   (25 / 5) = [(F – 32) / 9]

                   F = 77° F

Answer :

[(F – 32) / 9] = [(K – 273) / 5]

                   ∴ F – 32 = (9/5)(K – 273)

                   given : Temperature on kelvin scale = xk

                             Temperature on Fahrenheit = xF

                   ∴       F = K = x

                   ∴       x – 32 = (9/5)(x – 273)

                             (4/5)x = 459.67

                   ∴       x = 574.25   

Answer :

[(ΔT_c) / (100)] = [(ΔT_F) / (180)]

                   Temperature is lowered until Fahrenheit temperature is 140°C

                             ∴ Δ T_c = [(100) / (180)] × (212 – 140)

                             Δ T_c = 40°C i.e. fall in thermometer temperature 40°C

Answer :

Time period of pendulum = T = 2π√(ℓ/g)

                   as g is constant     T ∝ √ℓ

                                      ∴       T ∝ ℓ^1/2

                                                (ΔT / T) = (1/2)(Δℓ / ℓ)…….(1)

                             given:                   α = 2 × 10^–6

                                                Δt = 10°C

                                                (Δℓ / ℓ) = α ∙ Δt

                             from (1) = (ΔT / T) = (1/2) × (Δℓ / ℓ) = (1/2) × α × Δt

                                                          = (1/2) × 2 × 10^–6 × 10 = 10^–5

                                      [(ΔT / T) × 100] = 10^–5 × 10² = 10^–3

                             ∴       % increases in T = 1 × 10^–3 

Answer :

dV = V₂ – V₁ = 3 litre – 1 litre = 2 litre ∙ 2 × 10^–3 m³

                   pressure = p = 1 atm = 10⁵ pascal = 10⁵ N/m²

                   work done by gas = P ∙ dv

                                                = 2 × 10^–3 ×10⁵ = 200J