Thermodynamic quantity of heat?

Question

Vesel thermally insulated contain 2.1 kg H20 & 2.kgice. all at 0c the outlet of tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water how many grams of steam must becondense inside the vessel to raise the teperature of the system to 34 dgree celcius. neglect the theat transferred to the container

Answer 1

the final common temp in vessel is 34 C
let m kg steam condenses
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heat lost by steam >
* at 100 C (phase change) = m L(steam)
* from 100 C steam-water to 34 C water = m * c (w) * (100 - 34)
===========
heat gained by ice+cold water >
* 2kg ice to ice-water (phase change) = 2*L(ice)
* (2.1+2) kg ice-water from 0 C to 34 C water = 4.1*c (w) * (34-0)
==================
energy conservation at equilibrium
heat lost by steam = gained by ice=coldwater
m L(steam) + m*c(w)*(100 - 34) = 2*L(ice) + 4.1*c(w)*(34-0)
m [2272*10^3+4180*66] = 2*334*10^3 + 4.1*4180*34
m = 1250692 / 2547880
m = 490.8 gram steam will condense

Answer 2

You need to do a heat balance. I don't have numbers available, so let delHv = heat to vaporize 1 gram of water to steam at 100 deg C , and Cp= specific heat of water and
delHf= heat to freeze 1 gram of water at 0 deg C.
If Sw grams of steam are required, then:
Heat added to system from steam condensation and cooling:
Sw*delHv + Sw*Cp*(100-34)
Heat used by system:
2100*Cp*34 + 2000*Cp*34 + 2000*delHf.
You can set these equal to each other, and solve for Sw.