enthalpy

dH=Tds+Vdp

T and p are constant..

so dH=Tds

so it will be same as change of ds heat will be absorbed so it should be +ve and increase?

ΔH = ΔU + PΔV

with inc in temperature ΔU increases...
but in water, ΔV decreases till 4°C....

so may be, decrease of ΔV > increase of int energy...
so, overall effect is decrease in enthalpy...

Hey apporva
I think u r totally confused really

same question by u at:
TargetIIT » Forum » Thermal Physics » melting of ice 274K

there u have written that answer is it increases and sky there is saying it decreases

here u have given diff answer and here everyone is giving opposite.

I think Sky should answer here and all u there.

see,,,,
(our sir explained.....)

ΔH = ΔU + PΔV

ΔU=0 since no temp change as delU= nCvdelT

bt, del V decresaes in water till 4 degree cel.

so finally,,,, delH decreases.......

the above question is put in physics forum...(correctly)..:)