Question

One kilogram of water at 100 0C is cooled reversibly to 15 0C. Compute the change in entropy. Specific heat of water is 4190 J/Kg.K.

Answer 1

Answer:

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Explanation:

If the water is cooled reversibly with no phase changes, then there is no entropy generation during the entire process. By the Second Law of Thermodynamics, we represent the change of entropy ($s_{2} - s_{1}$), in joules per gram-Kelvin, by the following model:

$s_{2} - s_{1} = \int\limits^{T_{2}}_{T_{1}} {\frac{dQ}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \int\limits^{T_{2}}_{T_{1}} {\frac{dT}{T} }$

$s_{2} - s_{1} = m\cdot c_{w} \cdot \ln \frac{T_{2}}{T_{1}}$ (1)

Where:

$m$ - Mass, in kilograms.

$c_{w}$ - Specific heat of water, in joules per kilogram-Kelvin.

$T_{1}$, $T_{2}$ - Initial and final temperatures of water, in Kelvin.

If we know that $m = 1\,kg$, $c_{w} = 4190\,\frac{J}{kg\cdot K}$, $T_{1} = 373.15\,K$ and $T_{2} = 288.15\,K$, then the change in entropy for the entire process is:

$s_{2} - s_{1} = (1\,kg) \cdot \left(4190\,\frac{J}{kg\cdot K} \right)\cdot \ln \frac{288.15\,K}{373.15\,K}$

$s_{2} - s_{1} = -1083.112\,\frac{J}{kg\cdot K}$

The change in entropy is -1083.112 joules per kilogram-Kelvin.

Answer 2

Answer:

• The change in entropy = $-1083.534 J/k$

Explanation:

Change in entropy,

$\delta S = mCp * In[\frac{T2}{T1}]$

The initial temperature,

$T1 = 100^oC\\\\T1 = 100+273\\\\T1 = 373k$

Final value of temperature,

$T2 = 15^oC\\\\T2 = 15+273\\\\T2 = 288k$

where,

$m = 1kg\\\\Cp = 4190 J/kg.k$

Substitute into $\delta S$

$\delta S = mCp * In[\frac{T2}{T1}]\\\\\delta S = 1 * 4190 * In[\frac{288}{373}]\\\\\delta S = 4190 * In[0.7721]\\\\\delta S = 4190 * [-0.2586]\\\\\delta S = -1083.534 J/k$

The negative sign exists because the change in entropy will be decreasing due to cooling.

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