A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
If 19 g of steam at 100°C is injected into the bucket, what is the final equilibrium temperature of the system?
What mass of ice remains?

The final equilibrium temperature of the system is

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice = 129 g = 0.129 kg

Mass of the steam = 19 g = 0.019 kg

Initial temperature is   = 0°C

Temperature of  steam   = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

Where is a constant known as heat of fusion  and the value is

The energy been released when the steam changes to water is mathematically given as

           Where is a constant known as heat of vaporization and the value is

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 100°C)

Where is the specific heat of water which has a value

Looking at the values we obtained we noticed that ]

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.