A solid metallic block weighing 5 kg has an initial temperature of 500°C. 40 kg of water initially at 25°C is contained in a perfectly insulated tank. The metallic block is brought into contact with water. Both of them come to equilibrium. Specific heat of block material is 0.4 kJ.kg^-1. K^-1. Ignoring the effect of expansion and contraction and also the heat capacity to tank, the total entropy change in kJ.kg^-1 , K^-1 is

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 (Δ𝑆blockΔSblock​) can be calculated using: Δ𝑆block=𝑄block𝑇blockΔSblock​=Tblock​Qblock​​ Similarly, the change in entropy for the water (Δ𝑆waterΔSwater​) can be calculated using: Δ𝑆water=𝑄water𝑇waterΔSwater​=Twater​Qwater​​ The total entropy change is the sum of these individual entropy changes: Δ𝑆total=Δ𝑆block+Δ𝑆waterΔStotal​=ΔSblock​+ΔSwater​ First, let's find the heat exchange (𝑄Q) for both the block and the water. For the block: 𝑄block=𝑚block⋅𝑐block⋅Δ𝑇blockQblock​=mblock​⋅cblock​⋅ΔTblock​ Q_{\text{block}} = 5 \, \text{kg} \cdot 0.4 \, \text{kJ/kg} \cdot (500 - T_{\text{water}}) \, ^\circ \text{C} For the water: 𝑄water=𝑚water⋅𝑐water⋅Δ𝑇waterQwater​=mwater​⋅cwater​⋅ΔTwater​ 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 𝑇blockTblock​ and 𝑇waterTwater​ 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 𝑇waterTwater​ where the total entropy change becomes zero. After solving these equations, we'll get the value of 𝑇waterTwater​, 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}}} Δ𝑆total=Δ𝑆block+Δ𝑆waterΔStotal​=ΔSblock​+ΔSwater​ Δ𝑆total=2 (500−𝑇water)𝑇block+167 (𝑇water−25)𝑇waterΔStotal​=Tblock​2(500−Twater​)​+Twater​167(Twater​−25)​ Δ𝑆total=2 (500−𝑇water)⋅𝑇water+167 (𝑇water−25)⋅𝑇block𝑇block⋅𝑇waterΔStotal​=Tblock​⋅Twater​2(500−Twater​)⋅Twater​+167(Twater​−25)⋅Tblock​​ Δ𝑆total=1000 𝑇water−2 𝑇water2+167 𝑇water−4175𝑇block⋅𝑇waterΔStotal​=Tblock​⋅Twater​1000Twater​−2Twater2​+167Twater​−4175​ Δ𝑆total=−2 𝑇water2+1167 𝑇water−4175𝑇block⋅𝑇waterΔStotal​=Tblock​⋅Twater​−2Twater2​+1167Twater​−4175​ Δ𝑆total=−2 (𝑇water−25)(𝑇water−4175)5⋅𝑇waterΔStotal​=5⋅Twater​−2(Twater​−25)(Twater​−4175)​ Setting Δ𝑆total=0ΔStotal​=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: Δ𝑆total=−2 (25−25)(25−4175)5⋅25=0 kJ/kg⋅K−1ΔStotal​=5⋅25−2(25−25)(25−4175)​=0kJ/kg⋅K−1 So, the total entropy change is 0 kJ/kg⋅K−10kJ/kg⋅K−1​.