Question

What is the change in entropy when 2.5 mole of water is heated from 27 degree celsius to 87 degree celsius?

Answer 1

Hope this will help you

Answer 2

Answer: The entropy change is 1.91 J/K

Explanation:

To calculate the change in entropy for the given change in temperature, we use the equation:

$\Delta S=nC_p\ln \frac{T_2}{T_1}$

where,

$\Delta S$ = change in entropy = ?

n = number of moles of water = 2.5 moles

$C_p$ = heat capacity at constant pressure = 4.186 J/K.mol

$T_1$ = initial temperature = $27^oC=]27+273]K=300K$

$T_2$ = final temperature = $87^oC=]87+273]K=360K$

Putting values in above equation, we get:

$\Delta S=2.5\times 4.186\times \ln (\frac{360}{300})\\\\\Delta S=1.91J/K$

Hence, the entropy change is 1.91 J/K