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
What is the specific heat of the unknown solid when a 225 g sample is heated to 67°C and placed into a calorimeter with 25.6 g of water at 15.6°C, resulting in a final temperature of 53°C? 1.271 J/g°C