Question 210: Thermodynamics & KTG

Question

Water of mass $m$ gram is slowly heated to increase the temperature from $T_1$ to $T_\gamma$. The change in entropy of the water, given specific heat of water is $1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$, is :

Accepted Answer

To determine the change in entropy when water is heated from temperature $$T_1$$ to $$T_2$$ (here, $$T_\gamma$$ is equivalent to $$T_2$$), we use the definition of entropy change for a reversible process:
$$ \Delta S = \int_{T_1}^{T_2} \frac{dQ}{T} $$
Since the water is being heated slowly, the process is reversible, and the heat added is given by:
$$ dQ = mc\,dT $$
where:
Ã¢ÂÂ¢ $$m$$ is the mass of the water,
Ã¢ÂÂ¢ $$c$$ is the specific heat capacity (given as $$1\,\mathrm{J\,kg^{-1}\,K^{-1}}$$).
Substitute $$dQ$$ into the integral:
$$ \Delta S = \int_{T_1}^{T_2} \frac{mc\,dT}{T} = mc\, \int_{T_1}^{T_2} \frac{dT}{T}. $$
Since $$c = 1$$, the equation simplifies to:
$$ \Delta S = m \int_{T_1}^{T_2} \frac{dT}{T}. $$
Evaluating the integral, we have:
$$ \int_{T_1}^{T_2} \frac{dT}{T} = \ln{T}\Big|_{T_1}^{T_2} = \ln\left(\frac{T_2}{T_1}\right). $$
Thus, the change in entropy is:
$$ \Delta S = m \ln\left(\frac{T_2}{T_1}\right). $$
Comparing with the options provided, the correct answer is:
Option D: $$m \ln\left(\frac{T_2}{T_1}\right).$$

Answer

$\mathrm{m}\left(\mathrm{T}_2-\mathrm{T}_1\right)$

Answer

zero

Answer

$\mathrm{m} \ln \left(\frac{\mathrm{T}_1}{\mathrm{~T}_2}\right)$

Answer

$\mathrm{m} \ln \left(\frac{\mathrm{T}_2}{\mathrm{~T}_1}\right)$

Thermodynamics & KTG

Water of mass m gram is slowly heated to increase the temperature from T_1 to T_\gamma. The change in entropy of the water, given specific heat of water is 1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}, is :

Choose the correct answer: