A container of 1.5 kg of water with a specific heat of 4186 J/(kg·°C) is heated from 25°C to 95°C. The change in the entropy of the water is

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The change in entropy ($\Delta S$) for a substance when it is heated or cooled can be calculated using the formula $\Delta S = mC\ln\frac{T_f}{T_i}$, where $m$ is the mass of the substance, $C$ is the specific heat capacity, $T_f$ is the final temperature in Kelvin, and $T_i$ is the initial temperature in Kelvin. First, convert the temperatures from Celsius to Kelvin by adding 273.15 to each. So, $T_i = 25 + 273.15 = 298.15 K$ and $T_f = 95 + 273.15 = 368.15 K$. Then, plug the values into the formula:$$\Delta S = 1.5 \times 4186 \times \ln\left(\frac{368.15}{298.15}\right)$$$$\Delta S \approx 1.5 \times 4186 \times \ln\left(1.235\right)$$$$\Delta S \approx 1.5 \times 4186 \times 0.212$$$$\Delta S \approx 1300 J/K$$Therefore, the change in entropy of the water is approximately 1300 J/K.

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