You inadvertently leave a pan of water boiling away on thehot stove. You return just in time to see the last drop converted into steam. The pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas?

6.05

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8.15

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6.05 kJ/K

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Solution

The correct option is D 6.05 kJ/K
$△ S_{H_{2} O} = \frac{Q_{a b s o r p t i o n b y H_{2} O}}{T} Q = m L_{V} = ρ V L_{V} △ S_{H_{2} O} = \frac{ρ V L_{V}}{T} = \frac{(1 \frac{k g}{L}) (1 L) (2257 \frac{k J}{k g})}{373 K} = 6.05 k J / K$

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You inadvertently leave a pan of water boiling away on thehot stove. You return just in time to see the last drop converted into steam. The pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas?

The correct option is D 6.05 kJ/K△SH2O=QabsorptionbyH2OTQ=mLV=ρVLV△SH2O=ρVLVT=(1kgL)(1L)(2257kJkg)373K=6.05kJ/K

The correct option is D 6.05 kJ/K
$△ S_{H_{2} O} = \frac{Q_{a b s o r p t i o n b y H_{2} O}}{T} Q = m L_{V} = ρ V L_{V} △ S_{H_{2} O} = \frac{ρ V L_{V}}{T} = \frac{(1 \frac{k g}{L}) (1 L) (2257 \frac{k J}{k g})}{373 K} = 6.05 k J / K$