Physics

Heat, Temperature and Internal Energy

Physics·Revision Notes

Specific Heat — Revision Notes

NEET UG
Version 1Updated 24 Mar 2026

⚡ 30-Second Revision

  • Specific Heat ($c$)Heat to change 1,kg1,\text{kg} by 1,K1,\text{K}. Unit: J kg1^{-1} K1^{-1}.
  • FormulaQ=mcDeltaTQ = mcDelta T
  • Heat Capacity ($C$)Heat to change object by 1,K1,\text{K}. Unit: J K1^{-1}. C=mcC = mc.
  • Molar Specific Heat ($C_m$)Heat to change 1,mol1,\text{mol} by 1,K1,\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 Heatsgamma=Cp/Cvgamma = C_p / C_v.

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

  • Internal Energy ChangeDeltaU=nCvDeltaTDelta U = nC_vDelta T.
  • CalorimetryHeat Lost = Heat Gained (m1c1DeltaT1=m2c2DeltaT2m_1c_1Delta T_1 = m_2c_2Delta 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 gamma=Cp/Cvgamma = C_p/C_v varies with the gas's atomicity (degrees of freedom, ff). For monatomic gases, gamma=5/3gamma = 5/3; for diatomic gases at moderate temperatures, gamma=7/5gamma = 7/5. Internal energy change for an ideal gas is DeltaU=nCvDeltaTDelta 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 (DeltaTDelta T) of a unit mass (mm) by one degree. The core equation is Q=mcDeltaTQ = mcDelta T. Remember that DeltaTDelta 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 (DeltaU=nCvDeltaTDelta 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 gamma=Cp/Cvgamma = 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, gamma=53gamma = \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, gamma=75gamma = \frac{7}{5}.

Example: Calculate the heat required to raise the temperature of 0.5,kg0.5,\text{kg} of water from 20circC20^circ\text{C} to 70circC70^circ\text{C}. (Specific heat of water cw=4186,J kg1K1c_w = 4186,\text{J kg}^{-1}\text{K}^{-1}). m=0.5,kgm = 0.5,\text{kg}, c=4186,J kg1K1c = 4186,\text{J kg}^{-1}\text{K}^{-1}, DeltaT=7020=50circC=50,KDelta T = 70 - 20 = 50^circ\text{C} = 50,\text{K}. Q=mcDeltaT=0.5×4186×50=2093×50=104650,J=104.65,kJQ = 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 100,g100,\text{g} of iron (cFe=450,J kg1K1c_{Fe} = 450,\text{J kg}^{-1}\text{K}^{-1}) at 100circC100^circ\text{C} is dropped into 200,g200,\text{g} of water (cw=4186,J kg1K1c_w = 4186,\text{J kg}^{-1}\text{K}^{-1}) at 20circC20^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=882.2TfimpliesTfapprox24.08circC21244 = 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

    1
  1. DefinitionSpecific heat capacity (cc) is the heat energy required to raise the temperature of a unit mass of a substance by 1,K1,\text{K} or 1circC1^circ\text{C}.
  2. 2
  3. FormulaQ=mcDeltaTQ = mcDelta T, where QQ is heat, mm is mass, cc is specific heat, DeltaTDelta T is temperature change.
  4. 3
  5. UnitsSI unit is J kg1^{-1} K1^{-1}. Other common unit: cal g1^{-1} °C1^{-1}. Conversion: 1,cal=4.186,J1,\text{cal} = 4.186,\text{J}.
  6. 4
  7. Heat CapacityC=mcC = mc. Total heat for a given object to change temperature by 1,K1,\text{K}. Unit: J K1^{-1}.
  8. 5
  9. Molar Specific Heat ($C_m$)Heat required to raise 1,mol1,\text{mol} of substance by 1,K1,\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. 4186,J kg1K14186,\text{J kg}^{-1}\text{K}^{-1} or 1,cal g1°C11,\text{cal g}^{-1}\text{°C}^{-1}.
  12. 7
  13. Specific Heats of GasesNot unique, depends on process.

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

    1
  1. Mayer's Relation (for ideal gases)CpCv=RC_p - C_v = R, where RR is the universal gas constant (8.314,J mol1K18.314,\text{J mol}^{-1}\text{K}^{-1}). This implies Cp>CvC_p > C_v.
  2. 2
  3. Ratio of Specific Heats ($gamma$)gamma=Cp/Cv=1+2fgamma = 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, gamma=53approx1.67gamma = \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, gamma=75=1.4gamma = \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. gammagamma values are lower, e.g., gammaapprox1.33gamma approx 1.33 for non-linear.

    1
  1. Calorimetry PrincipleIn an isolated system, heat lost by hot bodies = heat gained by cold bodies. sum(mcDeltaT)lost=sum(mcDeltaT)gainedsum (mcDelta T)_{lost} = sum (mcDelta T)_{gained}.
  2. 2
  3. Dulong-Petit Law (for solids)For many solids at high temperatures, molar specific heat Cvapprox3Rapprox24.9,J mol1K1C_v approx 3R approx 24.9,\text{J mol}^{-1}\text{K}^{-1}.
  4. 3
  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.

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