the final temperature is

Question

A chunk of magnesium weighing 18.1 grams and originally at 98.92°C is dropped into an insulated cup containing 76.4 grams of water at 23.86°C.

Assuming that all of the heat is transferred to the water, the final temperature of the water is

°C.

Answer 1

specific heat for magnesium = 1.02 J/g/degree (from the internet)

specific heat for water = 4.184 J/g/degree

Heat lost by Mg = heat gained by H2O

heat lost by Mg = q = mC∆T

m = mass = 18.1 g

C = specific heat = 1.02 J/g/degree

∆T = change in temperature = 98.92 - Tf where Tf is the final temperature

heat lost by Mg = (18.1 g)(1.02 J/g/deg)(98.92-Tf)

Heat gained by water = q = mC∆T

heat gained by water = (76.4 g)(4.184 J/g/deg)(Tf-23.86)

(18.1 g)(1.02 J/g/deg)(98.92-Tf) = (76.4 g)(4.184 J/g/deg)(Tf-23.86)

1826 -18.46Tf = 319.7Tf - 7627

338.16Tf = 9453

Tf = 27.95ºC = final temperature of water