Question

An aluminum ball weighing 100 g is heated and dropped into water. The amount of heat transferred to water is 6.3 kJ. What could be the initial and final temperatures of the ball? (caluminum = 0.90 J/g°C)

Answer 1

Using the equation q=mCΔT, having q=6.3*10^3J, m=100g, and C=0.90J/g*C,

6300=100(0.9)ΔT, so ΔT = 7000*C.

Now, here is where I am iffy about, so bear with me. Assuming that once this heated material is dipped into water, and the water doesn't heat up, consider the standard temp for warm water to be 26.99*C, or 27*C. Have that to be the final temp of the material. If you do consider the equilibrium of media, where solid and liquid level out their difference in temps, then go for something like 34.5*C.

So now, with ΔT=Tfinal-Tinitial, set 7000=Tf-(whatever you are comfortable with) and solve for Tf.

I do not know if your assignment is based on reasonable assumptions or strict facts at that point, but here is an answer relying on assumptions.

I hope you find luck, as well as this is of consideration.