Calculating Copper Calorimeter Mass

The question asks us to find the mass of a copper calorimeter given the amount of heat it absorbs, the resulting temperature rise, and the specific heat capacity of copper. This is a classic problem involving the concept of specific heat capacity and heat transfer.

When an object absorbs heat, its temperature changes. The amount of heat absorbed (\(Q\)) is related to the mass of the object (\(m\)), its specific heat capacity (\(c\)), and the change in temperature (\(\Delta T\)) by the following formula:

\(Q = mc\Delta T\)

In this problem, we are given:

• Heat absorbed (\(Q\)) = 3.6 kJ
• Temperature rise (\(\Delta T\)) = 45°C
• Specific heat capacity of copper (\(c\)) = 0.4 Jg^-1K^-1

We need to find the mass (\(m\)) of the copper calorimeter in grams.

First, let's make sure all our units are consistent. The specific heat capacity is given in Jg^-1K^-1. This means mass should be in grams (g), heat in Joules (J), and temperature change in Kelvin (K).

• The heat absorbed is given in kilojoules (kJ). We need to convert this to Joules (J).
  1 kJ = 1000 J
  So, \(Q = 3.6 \text{ kJ} = 3.6 \times 1000 \text{ J} = 3600 \text{ J}\).
• The temperature rise is given in degrees Celsius (°C). A change in temperature in Celsius is the same as a change in temperature in Kelvin.
  So, \(\Delta T = 45\text{^\circ C} = 45 \text{ K}\).
• The specific heat capacity is already in the correct units: \(c = 0.4 \text{ Jg}^{-1}\text{K}^{-1}\).

Now we can rearrange the formula \(Q = mc\Delta T\) to solve for mass (\(m\)):

\(m = \frac{Q}{c\Delta T}\)

Substitute the values we have (with consistent units) into the formula:

\(m = \frac{3600 \text{ J}}{(0.4 \text{ Jg}^{-1}\text{K}^{-1})(45 \text{ K})}\)

\(m = \frac{3600 \text{ J}}{(0.4 \times 45) \text{ Jg}^{-1}}\)

Calculate the product in the denominator:

\(0.4 \times 45 = 18\)

Substitute this back into the equation for mass:

\(m = \frac{3600 \text{ J}}{18 \text{ Jg}^{-1}}\)

Now, perform the division. The units cancel out to give grams (J / (J/g) = J * g / J = g).

\(m = \frac{3600}{18} \text{ g}\)

\(m = 200 \text{ g}\)

Therefore, the mass of the copper calorimeter is 200 grams.

Let's summarize the given information and the calculated result:

Key Concepts for Heat Transfer Calculations

Understanding the definitions of specific heat capacity, heat absorbed, and temperature change is crucial for solving problems like this.

• Heat Absorbed (\(Q\)): This is the amount of thermal energy transferred to an object, causing its temperature to change or its state to change. In this case, it's the energy absorbed by the copper calorimeter.
• Temperature Change (\(\Delta T\)): This is the difference between the final and initial temperatures of the object. A positive \(\Delta T\) means the temperature increased (heat was absorbed), and a negative \(\Delta T\) means the temperature decreased (heat was released).
• Specific Heat Capacity (\(c\)): This is a physical property of a substance that represents the amount of heat required to raise the temperature of one unit of mass (like 1 gram or 1 kilogram) of the substance by one degree (like 1°C or 1 K). It's a measure of how much energy a substance can store as internal energy for a given change in temperature. Different materials have different specific heat capacities.

The formula \(Q = mc\Delta T\) only applies when the heat transfer causes a change in temperature and not a change in the state of matter (like melting or boiling).

Revision Table: Thermodynamics Formulas

Here are some related formulas often used in thermodynamics and heat transfer problems:

Additional Information on Calorimetry

Calorimetry is the science of measuring heat. A calorimeter is a device used for this measurement. The copper calorimeter mentioned in the problem is a common type of calorimeter.

• Calorimeters are often insulated to minimize heat exchange with the surroundings, ensuring that the measured heat transfer occurs primarily within the calorimeter system.
• When an experiment is conducted inside a calorimeter, the heat absorbed or released by the substance being studied is equal to the heat released or absorbed by the calorimeter and the liquid (usually water) it contains (assuming ideal conditions).
• In this specific problem, we only considered the heat absorbed by the copper calorimeter itself. In a more complex calorimetry problem, you might need to account for the heat absorbed by a liquid inside the calorimeter as well, using the specific heat capacity of the liquid.
• The principle of conservation of energy is fundamental to calorimetry. In an isolated system, the total heat lost by some components equals the total heat gained by others.