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NDA 2026
Exam
NDA
Year
2026
Subject
General Ability
Chapter
General Science

NDA 2026 · General Ability · General Science

A solid of mass m has temperature-dependent specific heat as C(T) =...

A solid of mass mm has temperature-dependent specific heat as C(T)=C0+αTC(T) = C_0 + \alpha T, where C0C_0 and α\alpha are constants. The solid is heated from T1T_1 to T2T_2. Which one of the following is the correct expression for quantity of heat (Q)(Q) of the solid mass?

Options

  1. AQ=mC0(T2−T1)Q = mC_0(T_2 - T_1)
  2. BQ=m(T2−T1)[C0+α(T1+T2)]Q = m(T_2 - T_1)[C_0 + \alpha(T_1 + T_2)]
  3. CQ=m(T2−T1)[C0(T1+T2)+α]Q = m(T_2 - T_1)[C_0(T_1 + T_2) + \alpha]
  4. DQ=m(T2−T1)[C0+α2(T1+T2)]Q = m(T_2 - T_1) \left[C_0 + \dfrac{\alpha}{2}(T_1 + T_2)\right]

Correct answer

D. Q=m(T2−T1)[C0+α2(T1+T2)]Q = m(T_2 - T_1) \left[C_0 + \dfrac{\alpha}{2}(T_1 + T_2)\right]

Step-by-step solution

The quantity of heat dQdQ required to raise the temperature of a mass mm by dTdT is given by:

dQ=mC(T)dTdQ = m C(T) dT

Substituting the given specific heat C(T)=C0+αTC(T) = C_0 + \alpha T:

dQ=m(C0+αT)dTdQ = m (C_0 + \alpha T) dT

Integrating both sides from T1T_1 to T2T_2:

Q=∫T1T2m(C0+αT)dTQ = \int_{T_1}^{T_2} m (C_0 + \alpha T) dT

Q=m[C0T+αT22]T1T2Q = m \left[ C_0 T + \dfrac{\alpha T^2}{2} \right]_{T_1}^{T_2}

Q=m[C0(T2−T1)+α2(T22−T12)]Q = m \left[C_0 (T_2 - T_1) + \dfrac{\alpha}{2} (T_2^2 - T_1^2) \right]

Using the identity T22−T12=(T2−T1)(T2+T1)T_2^2 - T_1^2 = (T_2 - T_1)(T_2 + T_1):

Q=m(T2−T1)[C0+α2(T1+T2)]Q = m (T_2 - T_1) \left[C_0 + \dfrac{\alpha}{2} (T_1 + T_2) \right]

Answer: $Q = m(T_2 - T_1) \left[C_0 + \dfrac{\alpha}{2}(T_1 + T_2)\right]
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