Question

A sample of 100 g $H_{2}O$ is slowly heated from ${27}^{o}C$ to ${87}^{o}C$. The change in entropy during heating is $xJ$. Then $100x$ is _____.

[Specific heat of water = 4200J/(kg.K)]

Answer 1

Entropy can be defined as a thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

When the temperature is increased the change in entropy can be calculated as

$\delta S=m{C}_{p\left(l\right)}\delta T/T$

Integrating from $T_1$ to $T_2$, we get

$\Delta S=m{C}_{P\left(l\right)}\ln \frac{T_2}{T_1}$

Given

$T_1={27}^{o}C=300K$

$T_2={87}^{o}C=360K$

Mass of water $=100g=0.1Kg$

Specific heat of water $=4200J/(Kg.K)$

$\Delta S=0.1\times 4200\times ln\frac{360}{300}$

$\Delta S=76.57J=x$

$100x=7657$