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At 1 atm pressure, 1 g of water having a volume of 1 cm^3 becomes 1671 cm^3 of steam when boiled. The heat of vaporization of water at 1 atm is 539 cal g^(-1) . What is the change in internal energy during the process? |
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Answer» 539 cal Q=mL= 1 X 539 = 539 cal `THEREFORE` Work DONE W=P `(V_v- V_l)` `= 1.013xx10^5 XX(1671-1)xx10^(-6)` J `=169.2/4.18` cal = 40.5 cal `therefore` Change in INTERNAL energy U=539 cal - 40.5 cal = 498.5 cal |
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