ANS:B - 0
To find the total entropy change when the solid metallic block is brought into contact with water until thermal equilibrium is reached, we can use the principle of entropy change for both the block and the water.
The change in entropy for the block (ΔđblockΔSblockâ) can be calculated using:
Δđblock=đblockđblockΔSblockâ=TblockâQblockââ
Similarly, the change in entropy for the water (ΔđwaterΔSwaterâ) can be calculated using:
Δđwater=đwaterđwaterΔSwaterâ=TwaterâQwaterââ
The total entropy change is the sum of these individual entropy changes:
Δđtotal=Δđblock+ΔđwaterΔStotalâ=ΔSblockâ+ΔSwaterâ
First, let's find the heat exchange (đQ) for both the block and the water.
For the block: đblock=đblock⋅đblock⋅ΔđblockQblockâ=mblockâ⋅cblockâ⋅ΔTblockâ Q_{\text{block}} = 5 \, \text{kg} \cdot 0.4 \, \text{kJ/kg} \cdot (500 - T_{\text{water}}) \, ^\circ \text{C}
For the water: đwater=đwater⋅đwater⋅ΔđwaterQwaterâ=mwaterâ⋅cwaterâ⋅ΔTwaterâ Q_{\text{water}} = 40 \, \text{kg} \cdot 4.18 \, \text{kJ/kg} \cdot (T_{\text{water}} - 25) \, ^\circ \text{C}
Since the system reaches thermal equilibrium, the final temperature đblockTblockâ and đwaterTwaterâ will be the same.
Now, let's calculate the entropy change for both the block and the water:
\Delta S_{\text{block}} = \frac{5 \, \text{kg} \cdot 0.4 \, \text{kJ/kg} \cdot (500 - T_{\text{water}}) \, ^\circ \text{C}}{T_{\text{block}}} \Delta S_{\text{water}} = \frac{40 \, \text{kg} \cdot 4.18 \, \text{kJ/kg} \cdot (T_{\text{water}} - 25) \, ^\circ \text{C}}{T_{\text{water}}}
Then, the total entropy change will be the sum of these two values.
Finally, we will find the value of đwaterTwaterâ where the total entropy change becomes zero.
After solving these equations, we'll get the value of đwaterTwaterâ, and by substituting it back into the total entropy change equation, we can find the total entropy change.
Let's calculate:
\Delta S_{\text{block}} = \frac{5 \, \text{kg} \cdot 0.4 \, \text{kJ/kg} \cdot (500 - T_{\text{water}}) \, ^\circ \text{C}}{T_{\text{block}}} = \frac{2 \, (500 - T_{\text{water}})}{T_{\text{block}}}
\Delta S_{\text{water}} = \frac{40 \, \text{kg} \cdot 4.18 \, \text{kJ/kg} \cdot (T_{\text{water}} - 25) \, ^\circ \text{C}}{T_{\text{water}}} = \frac{167 \, (T_{\text{water}} - 25)}{T_{\text{water}}}
Δđtotal=Δđblock+ΔđwaterΔStotalâ=ΔSblockâ+ΔSwaterâ Δđtotal=2 (500−đwater)đblock+167 (đwater−25)đwaterΔStotalâ=Tblockâ2(500−Twaterâ)â+Twaterâ167(Twaterâ−25)â
Δđtotal=2 (500−đwater)⋅đwater+167 (đwater−25)⋅đblockđblock⋅đwaterΔStotalâ=Tblockâ⋅Twaterâ2(500−Twaterâ)⋅Twaterâ+167(Twaterâ−25)⋅Tblockââ
Δđtotal=1000 đwater−2 đwater2+167 đwater−4175đblock⋅đwaterΔStotalâ=Tblockâ⋅Twaterâ1000Twaterâ−2Twater2â+167Twaterâ−4175â
Δđtotal=−2 đwater2+1167 đwater−4175đblock⋅đwaterΔStotalâ=Tblockâ⋅Twaterâ−2Twater2â+1167Twaterâ−4175â
Δđtotal=−2 (đwater−25)(đwater−4175)5⋅đwaterΔStotalâ=5⋅Twaterâ−2(Twaterâ−25)(Twaterâ−4175)â
Setting Δđtotal=0ΔStotalâ=0, we find T_{\text{water}} = 25 \, ^\circ \text{C} and T_{\text{water}} = 4175 \, ^\circ \text{C}. Since 4175 \, ^\circ \text{C} is not feasible, we discard it.
Therefore, T_{\text{water}} = 25 \, ^\circ \text{C}.
Now, substituting T_{\text{water}} = 25 \, ^\circ \text{C} into the total entropy change equation:
Δđtotal=−2 (25−25)(25−4175)5⋅25=0 kJ/kg⋅K−1ΔStotalâ=5⋅25−2(25−25)(25−4175)â=0kJ/kg⋅K−1
So, the total entropy change is 0 kJ/kg⋅K−10kJ/kg⋅K−1â.
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