Dropping a heated steel ball into ice and water

The second part of the question I'm doing asks 'What will happen if you drop a 100g steel ball at temperature 100c into a mixture of ice and water?'

so far I have

Dropping a 100˚C steel ball into a mixture of 500g ice and 500ml water at 0˚C
The temperature of the water will not change as the steel ball doesn’t have enough heat energy to melt all of the ice and increase the temperature of the water. All of the energy will be used in turning the solid ice into liquid water. This is the same reason that boiling water doesn’t exceed 100˚C.

Latent heat of melting for ice = 3.35x105J/KG
Q_ice=(m_ice)(L_m)

(0.5)(3.35x105) = 167500J

It would take 167500J to melt 0.5kg of ice

Im a little stuck with where to go next with it
thanks

Welcome to TSR
We would prefer it if you posted the whole question.
Especially as this is the 2nd part.

You need to find the amount of energy lost by the ball when it cools to zero to know if all the ice melts, or not.

thats the whole question i was given, there were no quantities of ice and water specified so i just made them up. I would have just explained that the temperature wont rise at all but then would it still be a quantitative answer?

Obviously what happens quantitatively depends on the quantity of ice and water in the mixture. Without a mark scheme or wider context to the question (eg book chapter theory if from a book; exam spec if from an exam) then you have to point out that in a mixture the temperature starts at zero and that initially only the ice melts so the heat from the ball goes into melting the ice. Then when and if it all melts (depending on quantities) the resulting mass of water plus melted ice will absorb heat from the ball and rise in temperature as you calculated in part one. You could assume quantities, as you have, and do that calculation.

I've had a couple of vague questions or questions that seem to have something missing on this access course so far but I checked with the teacher and he says the point is just to demonstrate that the temperature doesnt change before the ice is all gone. So if I do carry on assuming the quantities does

Dropping a 100g, 100˚C steel ball into a mixture of 500g ice and 500ml water at 0˚C

The temperature of the water will not change as the steel ball doesn’t have enough heat energy to melt all of the ice and increase the temperature of the water. All of the energy will be used in turning the solid ice into liquid water, only once all of the ice was melted would the heat energy start to raise the temperature of the water and melted ice.

Heat lost by steel ball = Heat gained by ice
Latent heat of melting for ice = 3.35x105J/KG
Q_ice=m_ice L_m
(0.5)(3.35x105) = 167500J
It would take 167500J to melt 0.5kg of ice

When the steel ball is dropped in the mixture and cools to 0˚C

Q=mc∆T
(0.1)(420)(100-0) = 4200J
mice = Qsteel / Lm = 4200/3.35x105

mice = 0.0125kg = 125g

Only 12.5g of ice would be melted when the steel ball is dropped into the mixture.


Thanks for the help