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GoodOldLimbo
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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.