The final temperature of the calorimeter was 17.7 o C. What was the initial temperature of the iron?

Question

A 26.0 g piece of iron (s = 0.46 J/g • ⁰C) is placed in a calorimeter with 200.0 g of water at 16.0 ⁰ C. The final temperature of the calorimeter was 17.7 ^o C. The specific heat of water is 4.184 J/g◦C What was the initial temperature of the iron?

Answer 1

The law of conservation of energy tells us that heat lost by iron must equal heat gained by water.

heat = q = mC∆T where m = mass; C = specific heat: ∆T = change in temperature

heat lost by iron = (26.0 g)(0.46 J/gº)(∆T)

heat gained by water = (200 g)(4.184 J/gº)(1.7º)

(26.0 g)(0.46 J/gº)(∆T) = (200 g)(4.184 J/gº)(1.7º)

11.96 ∆T = 1423

∆T = 119º This is the CHANGE in temperature of the iron.

Initial temperature of iron = final temperature + change = 17.7º + 119º = 136.7º (not corrected for sig. figs.)