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#unit conversion

6 public questions tagged with this topic.

Zeroth Law Thermal Equilibrium and Quasi-static

How many calories are equivalent to 2093 J of heat? (1 cal = 4.186 J )

**Thermal equilibrium** achieved when temperatures equal, zeroth law allows definition of temperature, quasi-static process approximates equilibrium at each step, enabling calculation of work as area under P-V curve, reversible processes are quasi-static without dissipative effects. Heat in cal = Heat in J4.186 . (2093)/(4.186) ≈ 500 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 500 cal, consistent with thermodynamic laws and energy conservation.

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

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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

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First Law of Thermodynamics Applications

How many joules are equivalent to 300 cal of heat? (1 cal = 4.186 J )

**First law of thermodynamics** ΔU = Q - W, ΔU internal energy change (J), Q heat added to system (J), W work done by system (J), sign convention physics Q positive when added, W positive when done by system, energy conservation, for isochoric W=0 ΔU=Q, for adiabatic Q=0 ΔU=-W, for isothermal ΔU=0 Q=W, for cyclic ΔU=0 Q_net=W_net. Heat in J = Heat in cal × 4.186 . 300 × 4.186 = 1255.8 J ≈ 1256 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

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

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SI Units and Dimensional Formulae

In a new unit system, 1unit of mass\=0.25kg, 1unit of length\=0.1m, 1unit of time\=0.5s. What is 3J (1J\=1kg m2s−2) in t

\[J\]=\[M L2T−2\]. 3J=3kg×(1m)2×(1s)−2=3×(4×0.25)×(10×0.1)2×(2×0.5)−2=3×1×1×4=12new units. As per NCERT Class 11 Chapter 2, dimensional analysis checks correctness and SI units define base and derived quantities. This principle confirms that 12 new is correct because its dimensions and unit match the physical quantity asked.

Ref: NCERT Class 11 Physics > Chapter 2: Units and Measurements > Topic: SI Units and Dimensional Formulae

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