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  • A sample of 100g water is slowly heated from <img alt="27^{\circ}C\: \: to\: \:87^{\circ}C" src="https://entrancecorner.oncodecogs.com/gif.latex?27%5E%7B%5Ccirc%7DC%5C%3A%20%5C%3A%20to%5C%3A%2

A sample of 100g water is slowly heated from 27^{\circ}C\: \: to\: \:87^{\circ}C. The change in entropy (in J/K) of the water is (specific heat capacity of water =4200\frac{J}{Kg-K})

Option: 1

76.6


Option: 2

86.6


Option: 3

56.6


Option: 4

66.6


Answers (1)

best_answer

Given:

T1 = 27oC = 273 + 27 = 300 K

T2 = 87oC = 273 + 87 = 360 K

m = 1000 g = 0.1 kg

Sw = 4200 J/kg-K

Now,

\\ \Delta Q = m \times s_w\times \Delta T \\ \\ \Delta S = \frac{\Delta Q}{T} \\ \\ \Delta S = \frac{ m \times s_w\times \Delta T}{T} \\ \\

\\ \\ \Rightarrow \int_{s1}^{s2}dS = \int_{T1}^{T2}ms_w\frac{dT}{T}\\ \\ \Rightarrow S_2-S_1=ms_w\ lnT|^{T_2}_{T_1} \\\\ \Rightarrow \Delta S=0.1 \times 4200 \times ln\frac{T_2}{T_1} \\\\ \Rightarrow \Delta S=0.1 \times 4200 \times ln\frac{360}{300} \\\\ \Rightarrow \Delta S =0.1 \times 4200 \times ln1.2 \\\\ \Rightarrow \Delta S =76.6JK^{-1}

Posted by

Divya Prakash Singh

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