Question

You have a calorimeter with 49.831 g of cool water (22.6 ℃ ). You add 30.704g of hot metal (98.2 ℃ ), and it all becomes 28.4 ℃ . Calorimeter constant is 82.5 J/ ℃ . What is the specific heat of metal?

Answer 1

When a hot metal is added to cool water, the temperature of the water rises and the temperature of the hot metal decreases to a common value.

Hence, the sum of heat gained by the cool water and the calorimeter must be equal to the heat lost by the hot metal.

Now, the temperature change for the cool water and calorimeter can be calculated as follows:

Initial temperature of cool water and calorimeter =

Final temperature of cool water and calorimeter =

Hence, the temperature change is

Hence, the total amount of heat absorbed by the cool water and calorimeter can be written as

Where

Hence, we can write

Hence, the amount of heat lost by the hot metal is 1687.76 J.

Now, the temperature change for the metal is

Given that mass of metal = 30.704 g.

We can write the equation for the heat lost by the metal as follows:

Hence, the specific heat of metal is approximately .