Question

An 50.04 g sample of aluminum is placed on a 25.84 g sample of copper initially at 97.16oC. If the heat is only transferred between the metals (with no loss to the surroundings) and the final temperature of both metals is 53.87oC, what is the inital temperature (in oC) of aluminum

Answer 1

final temperature is less than initial temperature of copper mean the energy going out of the copper is equal to the energy amount going into the aluminum. This means:

q_lost = q_gain

m1 * Cp1 * delta T = m2 * Cp2 * delta T

{ Cp of copper = 0.385 J/g.⁰C , Cp of aluminum = 0.902 J/g.⁰C }

So, by substitution, we then have:

(25.84 g) (97.16 ⁰C - 53.87 ⁰C) (0.385 J/g.⁰C) = (50.04 g) ( 53.87 ⁰C - X) (0.902 J/g.⁰C)

Solve for X

430.66 = 2431.48 - 45.136 X

X = 44.329 ⁰C