Question

You have a 28.2-g sample of a metal heated to 95.2°c. you drop it in a calorimeter with 100. g of water at 25.1°c. the final temperature of the water is 31.0°c. assuming no heat loss to the surroundings nor the calorimeter, calculate the heat capacity of the metal.

Answer 1

The heat lost by the metal should be equal to the heat gained by the water. We know that the heat capacity of water is simply 4.186 J / g °C. Therefore:

100 g * 4.186 J / g °C * (31°C – 25.1°C) = 28.2 g * Cp * (95.2°C - 31°C)

Cp = 1.36 J / g °C