This is chemistry

Question

A 25g piece of golf and a 25 g piece of aluminum, both heated to 100 degrees Celsius, are put in identical calorimeters. Each calorimeter contains 100.0 g of water at 20 degrees Celsius. What is the final temperature in the calorimeter containing the gold? In the calorimeter containing aluminum?

Answer 1

To answer this, we need the specific heats for gold, aluminum and water. Looking them up, I find

gold = 0.13 J/g/deg

aluminum = 0.91 J/g/deg

water = 4.184 J/g/deg

q = mC∆T where q = heat, m = mass, C = specific heat and ∆T = change in temperature

Heat lost by the metal = heat gained by the water, or put into an equation, we have

mC∆T_(metal) = mc∆T_(water)

For GOLD:

(25 g)(0.13 J/g/deg)(100 - Tf) = (100 g)(4.184 J/g/deg)(Tf - 20)

325 - 3.25Tf = 418.4Tf - 8368

421.65Tf = 8693

Tf = 20.6 degrees

For ALUMINUM:

(25g)(0.91 J/g/deg)(100 - Tf) = (100 g)(4.184 J/g/deg)(Tf - 20)

2275 - 22.75Tf = 418.4Tf - 8368

441.15Tf = 10643

Tf = 24.1 degrees