Generic solution

Given the following:

mi = 50 - Mass of ice (gm)
ti = 0 - Ice temperature (deg C)
ms = 10 - Mass of steam (gm)
ts = 100 - Steam temperature (deg C))

Using the following constants:

ci = 2100 - Ice thermal capacity (J/kg*deg)
l = 3.4*10^5 - Ice fusion heat (J/kg)
cw = 4100 - Water thermal capacity (J/kg*deg)
r = 2.2*10^6 - Steam condensing heat (J/kg)

Solution:

First of all, we need to figure out what will be inside at the end - just the ice, mixture of ice and water or just the water. Let's compare the amout of heat in the equation of thermal equiliblium in the first case. The heat of the condensing of all steam.

Q1 = r*ms*10^-3 = 2.2*10^4 J

Now, the heat to melt all the ice.

Q2 = l*mi*10^-3 + ci*mi*10^-3*(0 - ti) = 1.7*10^4 J

The head of the condensing is greater, then the ice will melt. The whole amount of heat of the condencing the steam and cooling the resulting water to 0 degree is

Q3 = Q1 + cw*ms*10^-3*(ts - 0) = 2.61*10^4 J

That is more than necessary to melt all the ice, thus to calculate the resulting temperuture let's write the equation.

Q2 + cw*mi*10^-3*(t - 0) = Q1 + cw*ms*10^-3*(ts - t)

t = (r*m*10^-3 + cw*ms*10^-3*ts + ci*mi*10^-3*ti - l*mi*10^-3) / (cw*(mi*10^-3 + ms*10^-3)) = 36.992 deg C

In the result we found that the whole amount of ice will be melted and the final temperature of water will be 37 degree celsius.