1ΔH = 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...
7 Answers
1dH=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?
1
33Hey 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.
1see,,,,
(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.......
