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:
Show Hint
- zero
- \( m (T_2 - T_1) \)
- \( m \ln \left( \frac{T_1}{T_2} \right) \)
- \( m \ln \left( \frac{T_2}{T_1} \right) \)
The Correct Option is D
Approach Solution - 1
To determine the change in entropy of water as its temperature increases from \( T_1 \) to \( T_2 \), we will use the concept of thermodynamic entropy change.
The change in entropy, \( \Delta S \), when an object is heated at constant pressure is given by the formula:
- Where:
- \( m \) is the mass of the substance in kilograms (since it's given in grams, convert it by dividing by 1000),
- \( c \) is the specific heat capacity of the substance, and
- \( T_1 \) and \( T_2 \) are the initial and final temperatures in Kelvin.
In this question, the specific heat of water is provided as \( 1 \, \text{J} \, \text{kg}^{-1} \, \text{K}^{-1} \). Therefore, the specific heat capacity \( c \) can be taken as 1 for our calculations.
Substituting the given values into the expression for entropy change, we get:
Thus, the change in entropy of the water is expressed as:
This matches the option: \(m \ln \left( \frac{T_2}{T_1} \right)\).
The other options can be ruled out based on incorrect application of the entropy formula or incorrect direction (such as zero change indicating no temperature change, which is not the case here).
Approach Solution -2
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