Question

CHEM 1405- Experiment 4: Matter and Energy 71 Roll # Name: Pre-lab Questions Section: Calculate the amout of heat water absorbs from a piece of hot metal using the following experimental data: 75.0 g cold water is taken in a calorimeter. The initial temperature of the water in the calorimeter is 21.2°C. To the calorimeter containing cold water 29.458 g metal at 98.9°C is added. The final temperature of the contents of the calorimeter is measured to be 29.5 °C. (Given: Specific heat of water 4.184 J. g °C) 1. 2. (Continuing from the previous question.) Knowing that the amount of heat the water absorbs (positive q) is equal but opposite to the heat the the metal lost (negative q), find the specific heat of the metal.

Answer 1

1.

Heat change,

q = m * s * ( t₂ - t₁ )

q = 75.0 * 4.184 * ( 29.5 - 21.2 )

q = 2604.54 J of heat is absorbed by water

q = 2.60 kJ

2.

Amount of heat absorbed by water = amount of heat lost by metal

2.60 = - m * s * ( t₂ - t₁)

2.60 = - 49.258 * s * ( 29.5 - 98.9)

2.60 = 3418.5052 * s

s = specific heat of metal = 0.0007606 kJ.g^-1.⁰C^-1 = 0.7606 J.g^-1.⁰C^-1