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songoku
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Homework Statement
The first law of thermodynamics may be expressed in the form: ΔU = q + w, where ΔU is the increase in internal energy, q is the thermal energy supplied to the system, w is the work done on the system.
Is ΔU, w and q positive, negative or zero when ice melts at 0oC to give water at 0oC (note: ice is less dense than water)?
Homework Equations
ΔU = q + w
ΔU = KE + PE
The Attempt at a Solution
When ice melts to give water at 0oC, the separation between molecules becomes smaller so the density of water will be bigger than ice. This will result in positive value of w.
No change in temperature means no change in KE of molecules but smaller separation between molecules means that the PE decreases so ΔU will be negative.
Based on first law of thermodynamics, the value of q should be negative because ΔU is negative and w is positive.
But the answer key says that ΔU and q are positive and w = 0.
Where did I go wrong? I am also confused why I got negative value of q. Ice turns to water requires heating so maybe q should be positive but it doesn't match the math.
Thanks
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