Answer to Question #73183 in Chemistry for Annie

Question

A 100.0g piece of iron is heated to a temperature of 125C and is placed in a bomb calorimeter containing 500mL of water at 10C. Calculate the final temperature of the water. The specific heat of water is 4.18J/gC and the specific heat of iron is 0.45J/gC.

Answer 1

Solution
Let’s take the final temperature of iron as 25C, so ΔT = 125 - 25 = 100C
Q1 = cm ΔT = 0.45J/gC x 100.0g x 100C = 4500J (for iron).
Q2 = cmΔT - for water. In the equilibrium Q1 = Q2, so
ΔT = 4500J / (4.18J/gC x 500g) = 2.15C (for water)
T2 = 10C + 2.15C = 12.15C
Answer: 12.15C.