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#heat supplied

3 public questions tagged with this topic.

Work Heat Distinction and Energy Transfer Modes

A solid of 2 moles is heated from 300 K to 320 K . If its molar specific heat capacity is 25.5 J mol⁻¹ K⁻¹ , what is the

**Work and heat** both energy transfer modes, work organized, e.g., lifting weight, compressing gas, electrical current, heat random due to temperature difference, work can be completely converted to heat via friction, but heat cannot be completely converted to work (second law), energy transfer modes include work (mechanical, electrical) and heat (conduction, convection, radiation). Δ Q = μ C Δ T . μ = 2 , C = 25.5 , Δ T = 320 - 300 = 20 . Δ Q = 2 × 25.5 × 20 = 1020 J . Using first law ΔU = Q - W, W = ∫ P dV, isobaric W

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

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Work Heat Distinction and Energy Transfer Modes

In an isobaric process, 1.2 moles of gas expand from 350 K to 420 K . What is the heat supplied if C_p = 25.5 J mol⁻¹ K⁻

**Modes of energy transfer** work is force times displacement, e.g., gas expansion W = P ΔV, heat is due to temperature gradient, internal energy change same for different combinations of Q and W, e.g., same ΔU can be achieved by adding heat at constant volume or doing work adiabatically, illustrating equivalence but distinction in mechanism. Δ Q = μ C_p Δ T . μ = 1.2 , C_p = 25.5 , Δ T = 420 - 350 = 70 . Δ Q = 1.2 × 25.5 × 70 = 2142 J . Using first law ΔU = Q - W, W = ∫ P dV,

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

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Specific Heat and Calorimetry

How much ice at 0∘C will melt if 16744J of heat is supplied? (Latent heat of fusion of ice = 3.35×105J kg−1)

Given: Q = 16744J, Lf = 3.35×105J kg−1. Q = mLf⇒m = QLf = 167443.35×105≈0.05kg = 50g. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 50 g. This satisfies dimensional consistency and physical conditions given, so option B is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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