[gsel]

10,00 g ice at 0,0°C is added to 100,0 g water as a fluid at 100,0°C. The reactions is adiabatic and the pressure assumed to be constant. What will be the end temperature of the water?

Melting enthalpy water is 5980J/mol at 0oC.
Heatcapacity og liquid water is 4,184 J/g K and independent of temperature.

[orgo814]

If I see some work, I may be able to guide you

[gsel]

qice  = n ∙ ΔfusH = 0.55506mol ∙ 5980J/mol = 3319.3J

qwater = n ∙ c ∙m ∙ ΔT = 5.5506mol ∙4.184 J/(g ∙K)  ∙100.0g∙〖(T〗_2- 373.15K)=([2322.4  (J ∙mol)/K ∙ T_(2 )]- 866592.8 J∙mol)   

qvap = -n ∙ ΔfusH = - (5.5506mol ∙ 5980 J/mol) = 33192.6J


I see this process consist of three steps:
1: Ice fase from solid to liquid.
2: Water fase from vapor to liquid.
3: Equilibrium T between ice water and hot water.
qice + qwater – qvap = 0

3319.3J + [2322.4 (J ∙mol)/K  ∙ T_(2 ) ]- 866592.8 J∙mol- 33192.6J=0

2322.4  (J ∙mol)/K  ∙ T_(2 )= 899785.4 J – 3319J

T2 = (The answer is supposed to be 356.8K)

[Enthalpy]

Is "fluid" supposed to mean "vapour" in English?

In any case, I doubt about the signs. And even more, about a heat of fusion multiplied by an amount of vapour.

[gsel]

Fluid was supposed to mean vapor, yes.
How about amount of vapor subtracted from heat of fusion multiplied by qwater?