Question

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 \( 1 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} \), is:

• A. zero
• B. \( m (T_2 - T_1) \)
• C. \( m \ln \left( \frac{T_1}{T_2} \right) \)
• D. \( m \ln \left( \frac{T_2}{T_1} \right) \)

Hint:

The change in entropy for heating or cooling a substance can be calculated using \( \Delta S = m \cdot c \cdot \ln \left( \frac{T_2}{T_1} \right) \), where \( m \) is the mass, \( c \) is the specific heat capacity, and \( T_1 \) and \( T_2 \) are the initial and final temperatures.

Answer 1

The Correct Option is D

The change in entropy \( \Delta S \) of a substance when its temperature changes is given by: \[ \Delta S = m \cdot c \cdot \ln \left( \frac{T_2}{T_1} \right), \] where: - \( m \) is the mass of the substance, - \( c \) is the specific heat capacity, - \( T_1 \) is the initial temperature, - \( T_2 \) is the final temperature. Given that the specific heat capacity of water is \( c = 1 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} \), the formula becomes: \[ \Delta S = m \cdot \ln \left( \frac{T_2}{T_1} \right). \] Thus, the change in entropy of the water is: \[ \boxed{m \ln \left( \frac{T_2}{T_1} \right)}. \]