Temperature Rise of Steel Ball in Water/Ice

[GoodOldLimbo]

Homework Statement

If a steel ball of mass 100 grams at a temperature of 100° is dropped into 1 litre of water at 20°C, what is the temperature rise? What would happen if it were dropped into a mixture of ice and water at 0°C?

Homework Equations

I assume that I'm required to use ΔQ = mcΔT. Also the general idea that heat lost by test sample = heat gained by water.

The Attempt at a Solution

At first I assumed:

0.1 x C x (100-t¹) = 1 x 4200 x (t²-20)

But I don't think I can approach that due to a lack of knowledge on the change in temperature:

Then I assumed if I was supposed to use (100-20) for both temperatures. Which ends up in C = 42000. But even then I'm not sure where to go. Is it possible that there isn't enough information given? Or am I encourage to find the specific heat capacity of steel from another source?

Any help would be appreciated.

 

[Doc Al]

What is the same is the final temperature of the steel and the water. And, yes, you need to look up the specific heat of steel. (Check your book.)