4. What mass of steam that is initially at 120 degrees Celsius is needed to warm 350 g of water and its 300 g aluminum container from 20 degrees Celsius to 50 degrees Celsius?

Let's start with this one.

This is a heat transfer problem where we have to equate the thermal energy lost by the steam to the thermal energy gained by the water and aluminum container.

There are 3 stages as the steam loses thermal energy. In the first stage, the steam is cooled to 120-(50-20)=90 Celsius.

The thermal energy freed is then

In the second stage, the steam is converted to water. To find the thermal energy removed, we use the latent heat of vaporization and

.

In the third stage, the temperature of the water is raised to 50 C. This frees an amount of thermal energy

Now, if we equate the thermal energy lost by the steam to the thermal energy gained by the water and aluminum, then:

Finding the mass and solving for m gives us

WHEW... check my solution. Easy to make a mistake in all that and I have left my physics get rusty. Now, you have a template for other calorimetry or latent heat problems similar to this one.