Respuesta :
Explanation:
The given data is as follows.
m = 10.0 kg = 10,000 g (as 1 kg = 1000 g)
Initial temp. of block 1, [tex]T_{1} = 100^{o}C[/tex] = (100 + 273) K = 373 K
Initial temp. of block 2, [tex]T_{2} = 0^{o}C[/tex] = (0 + 273) K = 273 K
So, heat released by block 1 = heat gained by block 2
[tex]mC \Delta T = mC \times \Delta T[/tex]
[tex]10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C[/tex]
[tex]T_{f} - 100^{o}C = 0^{o}C - T_{f}[/tex]
[tex]2T_{f} = 100^{o}C[/tex]
[tex]T_{f} = 50^{o}C[/tex]
Convert temperature into kelvin as (50 + 273) K = 323 K.
Also, we know that the relation between enthalpy and temperature change is as follows.
[tex]\Delta H = mC \Delta T[/tex]
= [tex]10000 g \times 0.385 J/K g \times 323 K[/tex]
= 1243550 J
or, = 1243.5 kJ
Now, calculate entropy change for block 1 as follows.
[tex]\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}[/tex]
= [tex]10000 g \times 0.385 J/K g \times ln \frac{323}{373}[/tex]
= [tex]10000 g \times 0.385 J/K g \times -0.143[/tex]
= -554.12 J/K
Now, entropy change for block 2 is as follows.
[tex]\Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}[/tex]
= [tex]10000 g \times 0.385 J/K g \times ln \frac{323}{273}[/tex]
= [tex]10000 g \times 0.385 J/K g \times 0.168[/tex]
= 647.49 J/K
Hence, total entropy will be sum of entropy change of both the blocks.
[tex]\Delta S_{total} = \Delta S_{1} + \Delta S_{2}[/tex]
= -554.12 J/K + 647.49 J/K
= 93.37 J/K
Thus, we can conclude that for the given reaction [tex]\Delta H[/tex] is 1243.5 kJ and [tex]\Delta S_{total}[/tex] is 93.37 J/K.
In thermodynamics, enthalpy is the total amount of the internal energy of the system. For the given data enthalpy is 1243.5 kJ and entropy is 93.37 J/K.
What is entropy?
Entropy is the measurable thermodynamic property that is the amount of the thermal energy of the system that is not available.
Given,
Mass of the object (m) = 10,000 g
The initial temperature [tex](\rm T_{1})[/tex] of the block 1 = 373 K
The initial temperature [tex](\rm T_{2})[/tex] of the block 1 = 273 K
The final temperature of the block by the heat released and gained by the block will be given as,
[tex]\begin{aligned} \rm mC\Delta T &= \rm mC\Delta T\\\\10000 \times 0.385 \times (\rm T_{\rm f} - 100)^{\circ}\rm C &= 10000 \times 0.385 \times (0- \rm T_{\rm f})^{\circ}\rm C\\\\\rm T_{f} &= 50^{\circ}\rm C\end{aligned}[/tex]
The final temperature is [tex]50^{\circ}\rm C[/tex] or 323 K.
The equation for the enthalpy and temperature is:
[tex]\begin{aligned} \rm \Delta H &= \rm mC\Delta T\\\\&= 10000 \times 0.385 \times 323\\\\&= 1243.5\;\rm kJ\end{aligned}[/tex]
Entropy change for block 1 is calculated as:
[tex]\begin{aligned}\rm \Delta S_{1} &= \rm m C ln \dfrac{T_{f}}{T_{i}}\\\\&= 10000\times 0.385 \times \rm ln\dfrac{323}{373}\\\\&= -554.12\;\rm J/K\end{aligned}[/tex]
Entropy change for block 2 is calculated as:
[tex]\begin{aligned}\rm \Delta S_{2} &= \rm m C ln \dfrac{T_{f}}{T_{i}}\\\\&= 10000\times 0.385 \times \rm ln\dfrac{323}{273}\\\\&= 647.49\;\rm J/K\end{aligned}[/tex]
Now, the total entropy of the system is calculated by adding the two entropies as,
[tex]\begin{aligned}\rm \Delta S_{total} &= \rm \Delta S_{1} + \Delta S_{2}\\\\&= -554.12 + 647.49\\\\&= 93.37\;\rm J/K\end{aligned}[/tex]
Therefore, for the given reaction the enthalpy is 1243.5 kJ and entropy is 93.37 J/K.
Learn more about enthalpy and entropy here:
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