Water of mass m gram is slowly heated to increase the temperature from $$T_{1}$$ to $$T_{2}$$ The change in entropy of the water, given specific heat of water is $$1Jkg^{-1}K^{-1}$$, is :

For reversible heating,

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

For water,

$$dQ=mcdT$$

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

Integrating from $$T_1\ to\ T_2$$,

$$ΔS=\int^{_{ }}\frac{mcdT}{T}$$

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

Given specific heat

$$c=1\ \text{J kg}^{-1}\text{K}^{-1}$$

Thus

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

Create a FREE account and get:

Predict your JEE Main percentile, rank & performance in seconds

Check Your Score Now

Ask our AI anything

AI can make mistakes. Please verify important information.

Download resources