#heat

Total Internal Reflection and Critical Angle

What is the primary optical phenomenon responsible for the mirage effect observed on hot roads?

**TIR application** requires n₁>n₂ and θ₁>C. Critical angle formula derived from Snell's law with θ₂=90°, n₁ sinC = n₂ sin90° = n₂, so sinC = n₂/n₁. For glass-air sinC=1/1.52, C≈41°, for water-air 48.6°, determining cutoff for transmission. The mirage effect on hot roads results from total internal reflection. Hot air near the ground has a lower refractive index than cooler air above, causing light from the sky to bend upward when the angle of incidence exceeds the critical angle, creating an illusion of water. Substituting values gives Total internal reflection, which matches expected image p

Ref: NCERT > Physics Book > Ray Optics > Total Internal Reflection and Critical Angle

View Details

Work Heat Distinction and Energy Transfer Modes

What distinguishes work from heat as a mode of energy transfer?

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

View Details

Work Heat Distinction and Energy Transfer Modes

How many joules are equivalent to 350 cal of heat? (1 cal = 4.186 J )

**Energy transfer** first law ΔU = Q - W, W includes P-V work, shaft work, electrical work, Q includes conduction Fourier law, convection, radiation Stefan-Boltzmann, distinction important because work is controllable, heat spontaneous from hot to cold, entropy associated with heat not work, explaining why heat engine efficiency

Ref: NCERT > Physics Book > Thermodynamics > Work Heat Distinction and Energy Transfer Modes

View Details

Heat Transfer Work Distinction and Internal Energy Change

In an isothermal process for an ideal gas, what happens to the internal energy?

**Heat and work distinction** heat is energy transfer due to temperature difference, random molecular motion, work is organized energy transfer due to macroscopic force, e.g., piston movement, both path functions depend on process, not state, internal energy U state function depends only on state (T for ideal gas), ΔU path independent, Q and W path dependent but Q-W = ΔU path independent. For an ideal gas, internal energy ( U ) depends only on temperature. In an isothermal process, temperature remains constant ( Δ T = 0 ), so Δ U = 0 . Using first law ΔU = Q - W, W = ∫

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

View Details

Heat Transfer Work Distinction and Internal Energy Change

Which thermodynamic process involves a constant temperature?

**Internal energy change** ΔU = Q - W, for same initial and final states ΔU same regardless of path, but Q and W differ, e.g., isothermal expansion from V₁ to V₂ ΔU=0 Q=W=n R T ln(V₂/V₁), adiabatic expansion between same volumes ΔU=-W, Q=0, different Q,W same ΔU if same T change, illustrating state vs path functions. An isothermal process is characterized by constant temperature ( T = constant ). For an ideal gas, this implies P V = constant , with heat exchange balancing work done. Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal

Ref: NCERT > Physics Book > Thermodynamics > Heat Transfer Work Distinction and Internal Energy Change

View Details

Isochoric Processes and Pressure-Temperature

A system releases 670 J of heat and has 230 J of work done on it. What is the change in internal energy?

**Isochoric work** zero because dV=0, so W=∫ P dV=0, internal energy change equals heat added, Q = n C_v ΔT, C_v molar specific heat at constant volume, for monatomic 3/2 R, for diatomic 5/2 R, temperature change directly from heat input. First Law: Δ Q = Δ U + Δ W . Δ Q = -670 (heat released), Δ W = -230 (work on system). -670 = Δ U - 230 ⇒ Δ U = -670 + 230 = -440 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

View Details

Isochoric Processes and Pressure-Temperature

A system releases 760 J of heat and performs 240 J of work. What is the change in internal energy?

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. First Law: Δ Q = Δ U + Δ W . Δ Q = -760 (heat released), Δ W = 240 (work by system). -760 = Δ U + 240 ⇒ Δ U = -760 - 240 = -1000 J . Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

View Details

Isochoric Processes and Pressure-Temperature

A system releases 870 J of heat and performs 330 J of work. What is the change in internal energy?

**Isochoric process** constant volume ΔV=0, work W=0, first law ΔU = Q, all heat goes to internal energy, P/T = constant from ideal gas law P V = n R T at constant V, pressure proportional to temperature, P₁/T₁ = P₂/T₂, e.g., heating gas in rigid container pressure rises proportionally to T. First Law: Δ Q = Δ U + Δ W . Δ Q = -870 (heat released), Δ W = 330 (work by system). -870 = Δ U + 330 ⇒ Δ U = -870 - 330 = -1200 J . Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > Isochoric Processes and Pressure-Temperature

View Details

Zeroth Law Thermal Equilibrium and Quasi-static

A system releases 550 J of heat and performs 200 J of work. What is the change in internal energy?

**Zeroth law of thermodynamics** if two systems A and B each in thermal equilibrium with third C, then A and B in equilibrium with each other, defines temperature as property that is same for systems in thermal equilibrium, basis for thermometer, temperature scale, thermal equilibrium means no net heat flow, same temperature. First Law: Δ Q = Δ U + Δ W . Δ Q = -550 (heat released), Δ W = 200 (work by system). -550 = Δ U + 200 ⇒ Δ U = -550 - 200 = -750 J . Using first law ΔU = Q - W, W = ∫ P

Ref: NCERT > Physics Book > Thermodynamics > Zeroth Law Thermal Equilibrium and Quasi-static

View Details

Second Law Heat Engines and Kelvin-Planck

A system releases 500 J of heat and has 200 J of work done on it. What is the change in internal energy?

**Carnot engine** reversible engine operating between T_h and T_c has maximum efficiency η =1 - T_c/T_h, T in kelvin, e.g., T_h=400 K T_c=300 K η=0.25, real engines less due to irreversibilities, second law defines direction of spontaneous processes and entropy increase. First Law: Δ Q = Δ U + Δ W . Δ Q = -500 J (heat released), Δ W = -200 J (work on system). -500 = Δ U - 200 ⇒ Δ U = -500 + 200 = -300 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W =

Ref: NCERT > Physics Book > Thermodynamics > Second Law Heat Engines and Kelvin-Planck

View Details

First Law of Thermodynamics Applications

How many joules are equivalent to 250 cal of heat? (1 cal = 4.186 J )

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Heat in J = Heat in cal × 4.186 . 250 × 4.186 = 1046.5 J ≈ 1047 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 1047

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

View Details

First Law of Thermodynamics Applications

How many calories are equivalent to 836 J of heat? (1 cal = 4.186 J )

**Internal energy** state function depends only on temperature for ideal gas, U = f/2 n R T, change ΔU = n C_v ΔT, first law connects heat, work, internal energy, for expansion work done by gas positive, compression work done on gas negative, heat added positive. Heat in cal = Heat in J4.186 . (836)/(4.186) ≈ 199.71 ≈ 200 cal . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W = P ΔV, isothermal W = n R T ln(V₂/V₁), adiabatic P V^γ = const and η = 1 - T_c/T_h, evaluation yields 200 cal, consistent with thermodynamic laws and energy conservation.

Ref: NCERT > Physics Book > Thermodynamics > First Law of Thermodynamics Applications

View Details