Question

One kilogram of water at $0^{\circ }\text{C}$ is heated to ${100}^{\circ }\text{C}$. Compute its change in entropy in $\left(\text{J/K}\right)$.

• A. $4200\times \ln \left(\frac{373}{273}\right)$
• B. $4200\ln 100$
• C. $84$
• D. Zero

Answer 1

The correct option is A $4200\times \ln \left(\frac{373}{273}\right)$

Change in entropy is given by,

$dS=\frac{dQ}{T}$

$\Rightarrow {\int }_{S_i}^{S_f}dS=\int \frac{dQ}{T}$

Here, there is rise in temperature of water,

$\Rightarrow dQ=msdT$

$\Rightarrow {\int }_{S_i}^{S_f}dS={\int }_{T_i}^{T_f}\frac{msdT}{T}$

$\Rightarrow {\int }_{S_i}^{S_f}dS=ms{\int }_{T_i}^{T_f}\frac{dT}{T}=ms[\ln T{]}_{T_i}^{T_f}$

$\Rightarrow S_f-S_i=\Delta S=ms\left(\ln \frac{T_f}{T_i}\right)$

Here, $m=1\text{kg}$

$s=4200\frac{\text{J}}{\text{kg}\text{K}}$

$T_i=0^{\circ }\text{C}=273\text{K}$
$T_f={100}^{\circ }\text{C}=373\text{K}$

$\Delta S=1\times 4200\times \ln \left(\frac{373}{273}\right)$

$\therefore \Delta S=4200\ln \left(\frac{373}{273}\right)$

Hence, option $\left(\text{A}\right)$ is correct.