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50 cal of heat should be supplied to take a system from the state A to the state B through the path ACB as shown in Figure. The quantity of heat to be supplied to take it from A to B via ADB. |
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Answer» process ACB: `Delta_(ACB)=DeltaW_(AC)+DeltaW_(CB)` `=50xx10^3xx(400-200)xx10^-6+0` `=10J=(10)/(4.2)=2.4cal` `DeltaQ=50cal` `DeltaQ=DeltaU+DeltaW=U_B-U_A+DeltaW` `50=(U_B-U_A)+2.4` `U_B-U_A=47.6cal` Process ADB: `DeltaW_(ADB)=DeltaW_(AD)+DeltaW_(DB)` `=0+155xx10^3xx(400-200)xx10^-6` `=31J=(31)/(4.2)=7.4cal` `DeltaQ=DeltaU+DeltaW=U_B-U_A+7.4` `=47.6+7.4` `=55cal` |
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