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What is the change in entropy when 2.5 mole of water is heated from \[27^\circ C\] to \[{87^\circ }C\]? Assume that the heat capacity is constant. (\[{C_{p,m}}\](\[{H_2}O\])=4.2 J/g-K, ln (1.2) =0.18)
A.16.6 J/K
B.9.0 J/K
C.34.02 J/K
D.1.89 J/K

Answer
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Hint: Entropy of a system can be defined as the thermal energy per unit temperature of simply defining entropy is that it is the degree of molecular disorder or randomness of a system. To the system that cannot be used or is basically unavailable for doing useful work.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
The absolute entropy of any system cannot be accurately calculated. Hence, we always refer to entropy as the change in entropy for a given system. There are different formulae that can be used to calculate the change in the entropy of a system, it is usually dependent on the data given to us. Let us now understand the data given to us:
1.Number of moles of water = 2.5 moles
Hence, mass of water sample = (molar mass) (number of moles) = (18) (2.5)
2. \[{T_1} = 27^\circ C = 300K\]
3. \[{T_2} = 87^\circ C = 360K\]
4. \[{C_{p,m}}({H_2}O) = 4.2\dfrac{J}{{g - K}}\]
Hence, in this case, we are going to use the following formula to calculate the change in entropy of the system:
\[\Delta S = m.{C_{p,m}}.\ln (\dfrac{{{T_2}}}{{{T_1}}})\]
Substituting the given values in the above formula, we get
$\Rightarrow$ \[\Delta S = (18)(2.5)(4.2)\ln \left( {\dfrac{{360}}{{300}}} \right)\]
$\Rightarrow$ \[\Delta S = \left( {18} \right)\left( {2.5} \right)\left( {4.2} \right)\left[ {\ln \left( {1.2} \right)} \right]\]
$\Rightarrow$ \[\Delta S = 34.0f\dfrac{J}{K}\]

Hence, Option C is the correct option

Note: In this particular situation, the value of the entropy of the system is positive. But there are certain systems where the value of the entropy is negative. Negative entropy means that something is becoming less disordered. In order for something to become less disordered, energy must be used.