[Blueberries116]

The problem is as follows:

A flask of negligible heat capacity is filled with a mix of [itex]100[/itex] grams of ice at [itex]0^{\circ}C[/itex], 150 grams of water at [itex]40^{\circ}C[/itex] and [itex]50[/itex] grams of steam at [itex]100^{\circ}C[/itex]. Calculate the final composition of the mixture.

The answer supposedly is 300 grams of liquid.

I don't know exactly how to get there.

[AWK]

A task similar to your first problem - you need to take into account the heat of condensation of steam, no specific heat of ice is needed.

[Borek]

Stepwise, a key to the success

Start with any two components, say ice and water. Ice will melt, yes? Does 150g of water at 40°C contain enough heat to melt all the ice? If it does, you will end with just water, and you can easily calculate the final temperature. If it doesn't, you will end with a mixture at 0°C and you should be able to calculate how much water/ice is present.

Then proceed further - if it was just water, check if it can absorb all energy from the condensation. If so - you have just water, and finding the final temperature is simple. If it was a mixture - choose any two components of different temperatures again and proceed exactly as you would do if the third component was absent.

[Blueberries116]

A task similar to your first problem - you need to take into account the heat of condensation of steam, no specific heat of ice is needed.

I'm still stuck with this problem.

For steam:

[itex](50)(540)=27000[/itex]

For ice:

[itex](100)(80)=8000[/itex]

For water:

[itex](150)(1)(40)=6000[/itex]

But what to do from here? 

[AWK]

(50)(540)=27000
 For ice:
(100)(80)=8000

27000>>8000
The whole ice will be melted.
Heat absorbed by ice and water to 100°C
(100·80+100·100·1)+150·60·1=27000 hence the whole water will boil together with condensed steam. For this sum, greater than 27000 temperature will be lower than 100°C.