Share This Tutorial

Views 14

AQA GCSE FOUNDATION PHYSICS - Specific Heat Capacity Calculations

Author Zak  |  Date 2024-10-25 17:36:24  |  Category Physics
Back Back

AQA GCSE Foundation Physics: Specific Heat Capacity Calculations

This tutorial will guide you through calculating changes in thermal energy using the specific heat capacity formula.

Understanding Specific Heat Capacity

The Formula

The relationship between these quantities is represented by the formula:

?E = m * c * ?T

Where:

Example Calculation

Question: A 500g block of copper is heated from 20°C to 60°C. Calculate the change in thermal energy. (Specific heat capacity of copper = 385 J/kg°C)

Solution:

  1. Convert mass to kilograms: 500g = 0.5 kg
  2. Calculate the temperature change: ?T = 60°C - 20°C = 40°C
  3. Apply the formula: ?E = m * c * ?T ?E = 0.5 kg * 385 J/kg°C * 40°C ?E = 7700 J

Answer: The change in thermal energy of the copper block is 7700 J.

Tips for Solving Problems

Practice Problems

  1. A 2 kg block of aluminum is heated from 15°C to 55°C. Calculate the change in thermal energy. (Specific heat capacity of aluminum = 900 J/kg°C)
  2. A 1.5 kg iron bar absorbs 3000 J of energy. If the initial temperature was 25°C, what is the final temperature? (Specific heat capacity of iron = 450 J/kg°C)
  3. A 100 g copper pan is heated from 20°C to 100°C. How much energy is required? (Specific heat capacity of copper = 385 J/kg°C)

By practicing these calculations, you will gain a deeper understanding of specific heat capacity and its importance in thermal energy transfer.