Physics·Revision Notes

Specific Heat — Revision Notes

NEET UG
Updated 24 Mar 2026

⚡ 30-Second Revision

  • Specific Heat ($c$)Heat to change 1kg1\,\text{kg} by 1K1\,\text{K}. Unit: J kg1^{-1} K1^{-1}.
  • FormulaQ=mcDeltaTQ = mcDelta T
  • Heat Capacity ($C$)Heat to change object by 1K1\,\text{K}. Unit: J K1^{-1}. C=mcC = mc.
  • Molar Specific Heat ($C_m$)Heat to change 1mol1\,\text{mol} by 1K1\,\text{K}. Unit: J mol1^{-1} K1^{-1}. Cm=McC_m = Mc.
  • GasesCvC_v (constant volume), CpC_p (constant pressure).
  • Mayer's RelationCpCv=RC_p - C_v = R (for ideal gas).
  • Ratio of Specific Heatsγ=Cp/Cv\gamma = C_p / C_v.

- Monatomic (f=3f=3): Cv=32RC_v = \frac{3}{2}R, Cp=52RC_p = \frac{5}{2}R, γ=53\gamma = \frac{5}{3}. - Diatomic (f=5f=5 at moderate T): Cv=52RC_v = \frac{5}{2}R, Cp=72RC_p = \frac{7}{2}R, γ=75\gamma = \frac{7}{5}.

  • Internal Energy ChangeΔU=nCvDeltaT\Delta U = nC_vDelta T.
  • CalorimetryHeat Lost = Heat Gained (m1c1ΔT1=m2c2ΔT2m_1c_1\Delta T_1 = m_2c_2\Delta T_2).

2-Minute Revision

Specific heat capacity (cc) measures a substance's thermal inertia, quantifying the heat needed to raise the temperature of a unit mass by one degree. Its formula is Q=mcDeltaTQ = mcDelta T, with SI units of J kg1^{-1} K1^{-1}. This differs from heat capacity (C=mcC=mc), which is for a specific object. Water has a notably high specific heat due to hydrogen bonding, making it an excellent thermal regulator.

For ideal gases, specific heat depends on the process: CvC_v (constant volume) and CpC_p (constant pressure). Mayer's relation, CpCv=RC_p - C_v = R, is crucial, showing CpC_p is greater because of work done during expansion at constant pressure.

The ratio γ=Cp/Cv\gamma = C_p/C_v varies with the gas's atomicity (degrees of freedom, ff). For monatomic gases, γ=5/3\gamma = 5/3; for diatomic gases at moderate temperatures, γ=7/5\gamma = 7/5. Internal energy change for an ideal gas is ΔU=nCvDeltaT\Delta U = nC_vDelta T.

Calorimetry problems apply the principle of heat lost equals heat gained in an isolated system, using Q=mcDeltaTQ = mcDelta T for each component.

5-Minute Revision

Specific heat (cc) is a material property defining how much heat energy (QQ) is required to change the temperature (ΔT\Delta T) of a unit mass (mm) by one degree. The core equation is Q=mcDeltaTQ = mcDelta T. Remember that ΔT\Delta T can be in Celsius or Kelvin, as the change is the same. Heat capacity (C=mcC = mc) refers to a specific object's ability to store heat, while molar specific heat (Cm=McC_m = Mc) is per mole, useful for gases.

For ideal gases, we distinguish between molar specific heat at constant volume (CvC_v) and constant pressure (CpC_p). At constant volume, all heat goes to increasing internal energy (ΔU=nCvDeltaT\Delta U = nC_vDelta T).

At constant pressure, some heat also performs work (W=PDeltaV=nRDeltaTW = PDelta V = nRDelta T), so CpC_p is always greater than CvC_v. Mayer's relation elegantly connects them: CpCv=RC_p - C_v = R. The ratio γ=Cp/Cv\gamma = C_p/C_v is vital, determined by the gas's degrees of freedom (ff).

For monatomic gases (f=3f=3), Cv=32RC_v = \frac{3}{2}R, Cp=52RC_p = \frac{5}{2}R, γ=53\gamma = \frac{5}{3}. For diatomic gases at moderate temperatures (f=5f=5), Cv=52RC_v = \frac{5}{2}R, Cp=72RC_p = \frac{7}{2}R, γ=75\gamma = \frac{7}{5}.

Example: Calculate the heat required to raise the temperature of 0.5kg0.5\,\text{kg} of water from 20C20^\circ\text{C} to 70C70^\circ\text{C}. (Specific heat of water cw=4186J kg1K1c_w = 4186\,\text{J kg}^{-1}\text{K}^{-1}). m=0.5kgm = 0.5\,\text{kg}, c=4186J kg1K1c = 4186\,\text{J kg}^{-1}\text{K}^{-1}, ΔT=7020=50C=50K\Delta T = 70 - 20 = 50^\circ\text{C} = 50\,\text{K}. Q=mcDeltaT=0.5×4186×50=2093×50=104650J=104.65kJQ = mcDelta T = 0.5 \times 4186 \times 50 = 2093 \times 50 = 104650\,\text{J} = 104.65\,\text{kJ}.

Calorimetry problems involve heat exchange: Qlost=QgainedQ_{lost} = Q_{gained}. If 100g100\,\text{g} of iron (cFe=450J kg1K1c_{Fe} = 450\,\text{J kg}^{-1}\text{K}^{-1}) at 100C100^\circ\text{C} is dropped into 200g200\,\text{g} of water (cw=4186J kg1K1c_w = 4186\,\text{J kg}^{-1}\text{K}^{-1}) at 20C20^\circ\text{C}, find the final temperature TfT_f.

0.1×450×(100Tf)=0.2×4186×(Tf20)0.1 \times 450 \times (100 - T_f) = 0.2 \times 4186 \times (T_f - 20) 45(100Tf)=837.2(Tf20)45 (100 - T_f) = 837.2 (T_f - 20) 450045Tf=837.2Tf167444500 - 45T_f = 837.2T_f - 16744 21244=882.2Tf    Tf24.08C21244 = 882.2T_f \implies T_f \approx 24.08^\circ\text{C}.

Remember to differentiate specific heat from latent heat (phase change without temperature change) and be mindful of units and conversions.

Prelims Revision Notes

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  1. DefinitionSpecific heat capacity (cc) is the heat energy required to raise the temperature of a unit mass of a substance by 1K1\,\text{K} or 1C1^\circ\text{C}.
  2. 2
  3. FormulaQ=mcDeltaTQ = mcDelta T, where QQ is heat, mm is mass, cc is specific heat, ΔT\Delta T is temperature change.
  4. 3
  5. UnitsSI unit is J kg1^{-1} K1^{-1}. Other common unit: cal g1^{-1} °C1^{-1}. Conversion: 1cal=4.186J1\,\text{cal} = 4.186\,\text{J}.
  6. 4
  7. Heat CapacityC=mcC = mc. Total heat for a given object to change temperature by 1K1\,\text{K}. Unit: J K1^{-1}.
  8. 5
  9. Molar Specific Heat ($C_m$)Heat required to raise 1mol1\,\text{mol} of substance by 1K1\,\text{K}. Unit: J mol1^{-1} K1^{-1}. Cm=McC_m = Mc, where MM is molar mass.
  10. 6
  11. Specific Heat of WaterExceptionally high, approx. 4186J kg1K14186\,\text{J kg}^{-1}\text{K}^{-1} or 1cal g1°C11\,\text{cal g}^{-1}\text{°C}^{-1}.
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  13. Specific Heats of GasesNot unique, depends on process.

* **Constant Volume (CvC_v)**: All heat increases internal energy. ΔU=nCvDeltaT\Delta U = nC_vDelta T. * **Constant Pressure (CpC_p)**: Heat increases internal energy and does work. Qp=ΔU+WQ_p = \Delta U + W.

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  1. Mayer's Relation (for ideal gases)CpCv=RC_p - C_v = R, where RR is the universal gas constant (8.314J mol1K18.314\,\text{J mol}^{-1}\text{K}^{-1}). This implies Cp>CvC_p > C_v.
  2. 2
  3. Ratio of Specific Heats ($\gamma$)γ=Cp/Cv=1+2f\gamma = C_p / C_v = 1 + \frac{2}{f}, where ff is degrees of freedom.

* Monatomic Gas (e.g., He, Ne): f=3f=3. Cv=32RC_v = \frac{3}{2}R, Cp=52RC_p = \frac{5}{2}R, γ=531.67\gamma = \frac{5}{3} \approx 1.67. * **Diatomic Gas (e.g., O2_2, N2_2)**: f=5f=5 (3 translational + 2 rotational at moderate T). Cv=52RC_v = \frac{5}{2}R, Cp=72RC_p = \frac{7}{2}R, γ=75=1.4\gamma = \frac{7}{5} = 1.4. * **Polyatomic Gas (e.g., CO2_2, NH3_3)**: f=6f=6 (non-linear) or f=5f=5 (linear) at moderate T. γ\gamma values are lower, e.g., γ1.33\gamma \approx 1.33 for non-linear.

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  1. Calorimetry PrincipleIn an isolated system, heat lost by hot bodies = heat gained by cold bodies. (mcDeltaT)lost=(mcDeltaT)gained\sum (mcDelta T)_{lost} = \sum (mcDelta T)_{gained}.
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  3. Dulong-Petit Law (for solids)For many solids at high temperatures, molar specific heat Cv3R24.9J mol1K1C_v \approx 3R \approx 24.9\,\text{J mol}^{-1}\text{K}^{-1}.
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  5. Distinction from Latent HeatSpecific heat involves temperature change without phase change. Latent heat involves phase change without temperature change.

Vyyuha Quick Recall

For Specific Heat, remember 'Q = MCAT' (pronounced 'Q equals M-Cat').

  • QHeat energy
  • MMass
  • CSpecific Heat Capacity
  • $\Delta$TChange in Temperature

This helps recall the primary formula. For gases, remember 'Cp is Greater than Cv by R' (Mayer's relation: CpCv=RC_p - C_v = R) because at constant pressure, the gas does 'R' amount of work per mole per Kelvin.