1. Entropy change for slow heating

For a substance heated reversibly from temperature $T_1$ to $T_2$, the entropy change is

$dS = \frac{dQ_{\text{rev}}}{T}$

Since the specific heat is constant,

$dQ = mc\,dT$

So,

$dS = \frac{mc\,dT}{T}$

Integrating from $T_1$ to $T_2$,

$\Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc \int_{T_1}^{T_2} \frac{dT}{T}$

$\Delta S = mc\ln\left(\frac{T_2}{T_1}\right)$

2. Given specific heat

The question gives specific heat of water as $c=1$ (though units appear misprinted; effectively take $c=1$).

Hence,

$\Delta S = m\ln\left(\frac{T_2}{T_1}\right)$

3. Check options

• A: $m(T_2-T_1)$ — this is not entropy.
• B: zero — incorrect, entropy increases on heating.
• C: $m\ln\left(\frac{T_1}{T_2}\right)$ — wrong sign.
• D: $m\ln\left(\frac{T_2}{T_1}\right)$ — correct.

Therefore, the correct option is D.