Thermodynamics: I can't solve this simple entropy exercise....

[babaliaris]

The text says:
"Steel bullet of 25kg with a Temperature of 400 Celsius, is being dropped on the bottom of an oil liquid of 100kg at a temperature of 100 Celsius. The system is isolated. Calculate
a) The change of entropy of the bullet,
b) the change of entropy of the oil,
c) the total change of entropy of the system."

also these values are given: Bullet $C_{p_{b}} = 0.5$ and oil $C_{p_{o}} = 2.5$

From the reading I've done so far in thermodynamics, my mind goes to these formulas for Isobaric:
$Q = nC_{p}ΔT$ and $ΔS = \frac{Q}{ΔT}$

The problem is that I don't know how to calculate the final Temperature of the oil and the bullet. I believe that if I know the final temperature of these two objects, then I can calculate the heat Q of them both individually and thous calculate the ΔS individually. But how do I do that? Probably I'm missing something in my notes. Also I'm not sure if this formula is correct: $ΔS = \frac{Q}{ΔT}$

Thank you

 

[Chestermiller]

You are correct. The first step is to determine the final temperature using the first law of thermodynamics. Do you know how to do that? You should, given that you are now learning about entropy and the second law of thermodynamics.

I have prepared a simple cookbook recipe on the step-by-step procedure you use to determine the change in entropy for a system like this: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/ There are worked examples in the cookbook. Please look it over and get back to me with questions.