Calculating the Specific Heat of an Unknown Solid

Introduction

When a sample of an unknown solid is heated and placed into a calorimeter with water, the final temperature of the system can be used to determine the specific heat of the solid. In this case, we have a 225 g sample of an unknown solid that is heated to 67°C and then placed into a calorimeter containing 25.6 g of water at 15.6°C. The final temperature of the system is 53°C. By applying the principle of conservation of energy, we can calculate the specific heat of the solid.

Calculation

Given data:

Mass of solid (m): 225 g

Initial temperature of solid (T1): 67°C

Final temperature of solid and water (T2): 53°C

Mass of water (m): 25.6 g

Initial temperature of water (T1): 15.6°C

Specific heat capacity of water (c): 4.18 J/g°C

Answer:

Specific heat of the solid = 1.271 J/g°C

Explanation:

Heat released by the metal sample will be equivalent to the heat absorbed by water.

By using the formula Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the temperature change, we can calculate the specific heat of the solid.

Heat released by the solid:

Q = 225 g × c × (67 - 53) = 3150 c joules

Heat absorbed by water:

Q = 25.6 g × 4.18 J/g°C × (53 - 15.6) = 4002.0992 joules

Setting the two heat values equal to each other, we get:

3150 c joules = 4002.0992 joules

c = 4002.0992 / 3150 = 1.271 J/g°C