Water of mass m gram is slowly heated to increase the temperature from T₁ to T_γ. The change in entropy of the water, given specific heat of water is 1 Jkg⁻¹ K⁻¹, is :

Solution

Physics sir · 4 steps

Temperature keeps changing, so you cannot just divide Q by T: add up dQ/T over every small step.

1. Small heat step. Slow heating is reversible, so dS = dQ/T. To raise the water by dT it absorbs dQ = mc dT, so dS = (mc dT)/T.

   Δ S = ∫_(T₁)^(T₂) (mc dT)/T = mc ln T₂/T₁

2. Put in c. With c = 1, this becomes Δ S = m ln T₂/T₁. The options use m directly, so m is taken in the same mass unit as the specific heat. The final temperature written as T_γ is T₂.
3. Check the sign. Since T₂ > T₁, the ratio is greater than 1 and ln T₂/T₁ is positive. Water that absorbs heat must gain entropy, so the sign is right.
4. So. The change in entropy is m ln T₂/T₁, which is option (d).

The idea to keep. When temperature changes during heating, integrate dQ/T. For a body of constant specific heat this always gives Δ S = mc ln T₂/T₁.

Answer from the NTA final answer key. Solution by PrepWiser.