Question

100gm of water is slowly heated from $ {27^ \circ }C $ to $ {87^ \circ }C $ . Calculate the change in the entropy of water (specific heat capacity of water $ = 4200{\text{J/kg - K}} $ )
(A) $ 40{\text{J/K}} $
(B) $ {\text{76}}{\text{.6J/K}} $
(C) $ {\text{100J/K}} $
(D) $ {\text{28J/K}} $

Answer 1

Hint: To solve this question, we need to use the basic expression for the change in entropy in a process. Then we have to substitute the expression for the heat exchange in terms of the change in temperature. On integrating the expression thus formed between the temperature limits given in the question, we will get the final answer.

Formula used: The formulae used for solving this question is given by
 $ dq = mcdT $ , here $ dq $ is the heat supplied to a substance of mass $ m $ and having a specific heat capacity of $ C $ , which is subjected to a change in temperature of $ dT $ .
 $ dS = \dfrac{{dq}}{T} $ , here $ dS $ is the change in entropy due to a heat exchange of $ dq $ at the temperature of $ T $ .

Complete step by step solution:
We know that the change in entropy is given by the relation
 $ dS = \dfrac{{dq}}{T} $ ……………….(1)
Also, the heat exchange is given in terms of the specific heat capacity by the relation
 $ dq = mcdT $ ……………….(2)
Putting (2) in (1) we get
 $ dS = \dfrac{{msdT}}{T} $ ……………….(3)
Now, according to the question, the water is heated from the temperature of $ {27^ \circ }C $ to $ {87^ \circ }C $ . Converting these temperature values to Kelvin scale, we get
 $ {T_1} = 27 + 273 = 300K $
 $ {T_2} = 87 + 273 = 360K $
Integrating (3) both sides between temperature limits of $ 300K $ and $ 360K $ , we get
 $ \int\limits_0^{\Delta S} {dS} = \int\limits_{300}^{360} {\dfrac{{msdT}}{T}} $
 $ \Rightarrow \left[ S \right]_0^{\Delta S} = ms\int\limits_{300}^{360} {\dfrac{{dT}}{T}} $
We know that $ \int {\dfrac{1}{x}dx = \ln x} $ . So we have
 $ \left[ S \right]_0^{\Delta S} = ms\left[ {\ln T} \right]_{300}^{360} $
On substituting the limits, we have
 $ \Delta S - 0 = ms\left( {\ln 360 - \ln 300} \right) $
Now, we know that $ \ln A - \ln B = \ln \left( {\dfrac{A}{B}} \right) $ . So we have
 $ \Delta S = ms\ln \left( {\dfrac{{360}}{{300}}} \right) $
According to the question, the mass of water is $ 100gm $ and its specific heat is equal to $ 4200{\text{J/kg - K}} $ . Therefore, we substitute $ m = 100g = 0.1kg $ and $ s = 4200{\text{J/kg - K}} $ in the above expression to get
 $ \Delta S = 0.1 \times 4200 \times \ln \left( {\dfrac{{360}}{{300}}} \right) $
On solving we get
 $ \Delta S = {\text{76}}{\text{.6J/K}} $
Thus, the value of the change in entropy of water is equal to $ {\text{76}}{\text{.6J/K}} $ .
Hence, the correct answer is option B.

Note:
In this question, the value of the mass of water is given to be equal to $ 100 $ grams. Also, the temperatures are given in degrees Celsius. But these are not the SI units. So do not forget to convert them into kilograms and Kelvin respectively, as done in the above solution.