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## Homework Statement

The first law of thermodynamics may be expressed in the form: Δ

*U*=

*q*+

*w,*where Δ

*U*is the increase in internal energy,

*q*is the thermal energy supplied to the system,

*w*is the work done on the system.

Is Δ

*U, w*and

*q*positive, negative or zero when ice melts at 0

^{o}C to give water at 0

^{o}C (note: ice is less dense than water)?

## Homework Equations

Δ

*U*=

*q*+

*w*

Δ

Δ

*U = KE + PE*## The Attempt at a Solution

When ice melts to give water at 0

^{o}C, 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 Δ

*U*will be negative.

Based on first law of thermodynamics, the value of

*q*should be negative because Δ

*U*is negative and

*w*is positive.

But the answer key says that Δ

*U*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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