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Let 1.00 Kg of liquid water at 100...

Let 1.00 Kg of liquid water at `100^@ C ` converted to steam at `100^@ C ` by boiling at standard atmospheric pressure ( which is 1.00 atm or `1.01 xx 10^5` Pa ) in the arrangement of Fig 21-7 the volume of that water changes from an initial value of `1.00 xx 10^3 m^3 ` as a liquid to `1.671 m^3 ` as steam
(c ) what is the change in the system internal energy during the process ?

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To solve the problem of finding the change in internal energy during the conversion of 1.00 kg of liquid water at 100°C to steam at 100°C, we will use the first law of thermodynamics. The first law states: \[ \Delta Q = \Delta U + W \] Where: - \(\Delta Q\) is the heat added to the system, ...
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