Question

Find the change in entropy when 3.0 kg of water is heated from 20°C to 80°C?

Answer 1

The change in entropy

$$\Delta S=\int_{T_1}^{T_2}\frac{dQ}{T}=\int_{T_1}^{T_2}\frac{cmdT}{T}$$

$$\Delta S=cm\ln\frac{T_2}{T_1}\\ =4200*3.0*\ln\frac{373}{293}=1321\:\rm J/ K$$