if we melt ice at 0'C to water at 0'C,is there any work done?

Work done is given by

Sp. heat at const pressure * change in temp

Since there is no change in temp, work done = zero.

The heat required to convert ice to water at constant temperature is called Latent (hidden) heat of melting of ice. This heat is ABSORBED but not used in rising the temperature. It is just utilized in increasing the internal energy of the ice. Hence both Q and del U are positive, but W is zero.

The density of ice is lesser than that of water. Hence, as the mass is constant, the volume decreases. This has NO connection with the work done.

we would desire to remember some thermodynamically data in this communicate. a million. below established rigidity situations water melts and freezes precisely at 0 °C, no longer above or under. 2. the two the liquid and the forged section can exist concurrently at this equilibrium temperature. 3. Spontaneously does no longer mean that the section substitute would not elect time. There must be warmth substitute between the water ranges. Spontaneously ability that if we've warmth pass the approach will bypass interior the path of extra suitable entropy of the full gadget. 4. stable water includes crystals. The crystal shape is powerful under 0 °C and at greater temperatures the crystal bonds get destroyed via thermal flow. to beat the bond capability you may desire to pass warmth. 5. a distinctive difficulty we've with "sub-cooled water", very organic water in liquid state under 0°C that could exist after careful cooling with none mechanical noise. This water can freeze on the instant through fact the warmth of section substitute may well be taken from the liquid water which has a temperature under equilibrium temperature of 0 °C.

In my thinking, work means the transfer of energy. Energy is needed to go into the system to break up all of the crystal structure of the ice. So I'd say that work has been done.

Interestingly, nobody else agrees with me so we will have to mark this question as interesting until somebody comes up with a better idea.

Yes, you are right,

in process of melting ice at standard atmospheric pressure P = 101 kPa, the ambient pressure does work

Work = -PΔV = 101 kPa x .0917 liter/kg = 9.26 Joule/kg

The change of internal energy E of water in transition from solid to liquid state is

ΔE = ΔQ + Work = ΔQ - PΔV = 333,550 J/kg + 9.26 J/kg

Note that mechanical work is tiny compared to heat ΔQ, that is why in high school physics this work is ignored.

The formula above can be rewritten as

ΔE + PΔV = ΔQ

Δ(E + PV) = ΔQ

The quantity is parenthesis (E + PV) is specal thermodynamic potential, called "enthalpy" and denoted by letter H:

H = E + PV

Heat of fussion quoted in ALL tables without execption is NOT change of internal energy E, but actual amount of HEAT ONLY ΔQ , equal to change of entalpy H. The difference becomes very important in process of evaporation, where expansion of volume is much grater, and work is no longer negligible. Or if ice is melted unger high pressures of thousands of atm, like on the bottom of ocean.

To summarize:

Heat required to melt 1kg of ice at 1 atm pressure is

ΔH = 333,550 J according to tables.

Increase of internal energy E is

ΔE = 333,559.26 J

due to extra mechanical work.

No work is done.

that needs more information... but virtually, none