Thermal equilibrium

Question

A chunk of aluminum weighing 18.5 grams and originally at 98.95 °C is dropped into an insulated cup containing 77.6 grams of water at 21.43 °C.

Assuming that all of the heat is transferred to the water, the final temperature of the water is BLANK  °C.

Answer 1

This problem is a little more difficult than the others you have posted, so here is a little bit of help. Hopefully you'll understand the concept.

HEAT LOST BY HOT ALUMINUM MUST EQUAL HEAT GAINED BY COLD WATER !!!

heat lost by Al = q = mC∆T = (18.5 g)(specific heat of Al)(98.95 - Tf) you'll have to look up the C of aluminum

heat gained by water = q = mC∆T = (77.6 g)(4.184 J/gº)(Tf - 21.43) the C for water is well known so I put it in

set the two equations equal to each other and after you look up and insert the specific heat for Al you have..

(18.5 g)(specific heat of Al)(98.95 - Tf) = (77.6 g)(4.184 J/gº)(Tf - 21.43) where Tf is the final temperature

All you have to do then is to do the math and solve for Tf which will be your answer.