SATHEE: Thermodynamics and Thermochemistry

Thermodynamics and Thermochemistry - Result Question 89

####14. Two blocks of the same metal having same mass and at temperature $T_{1}$ and $T_{2}$ respectively, are brought in contact with each other and allowed to attain thermal equilibrium at constant pressure. The change in entropy, $Δ S$ , for this process is

(2019 Main, 11 Jan I)

(a) $^{2} C_{p} \ln [\frac{{(T_{1} + T_{2})}^{1 / 2}}{T_{1} T_{2}}]$

(b) $^{2} C_{p} \ln [\frac{T_{1} + T_{2}}{4 T_{1} T_{2}}]$

(d) $^{2} C_{p} \ln [\frac{T_{1} + T_{2}}{2 T_{1} T_{2}}]$

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Solution:

At the thermal equilibrium,

final temperature $T_{f} = \frac{T_{1} + T_{2}}{2}$ $\Rightarrow$ for the 1st block, $Δ S_{I} = C_{p} \ln \frac{T_{f}}{T_{1}}$

$\Rightarrow$ for the 2nd block, $Δ S_{II} = C_{p} \ln \frac{T_{f}}{T_{2}}$

When brought in contact with each other,

$\begin{aligned} Δ S & = Δ S_{I} + Δ S_{II} = C_{p} \ln \frac{T_{f}}{T_{1}} + C_{p} \ln \frac{T_{f}}{T_{2}} & = C_{p} \ln (\frac{T_{f}}{T_{1}} \times \frac{T_{f}}{T_{2}}) = C_{p} \ln [\frac{T_{f}^{2}}{T_{1} T_{2}}] & = C_{p} \ln [\frac{{(\frac{T_{1} + T_{2}}{2})}^{2}}{T_{1} T_{2}}] = C_{p} \ln [\frac{{(T_{1} + T_{2})}^{2}}{4 T_{1} T_{2}}] \end{aligned}$

Thermodynamics and Thermochemistry - Result Question 89

Solution:

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Thermodynamics and Thermochemistry - Result Question 89

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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