Question

If it takes \(4.5 \mathrm{J}\) of energy to warm \(5.0 \mathrm{g}\) of aluminum from \(25^{\circ} \mathrm{C}\) to a certain higher temperature, then it will take ___ J to warm \(10 . \mathrm{g}\) of aluminum over the same temperature interval.

Short answer

It will take \(9\text{ J}\) to warm \(10\text{g}\) of aluminum over the same temperature interval.

Step-by-step solution

Write down the formula

We will use the heat transfer formula: \[ Q = mc\Delta T \]

Set up the proportion

We are given that 4.5 J will warm 5g of aluminum from \(25^{\circ} \mathrm{C}\) to another temperature, and we need to find the energy required to warm 10g of aluminum over the same temperature interval. Let's set up a proportion where we will compare the energy required to heat the masses of 5g and 10g of aluminum: \[ \frac{Q_1}{5\text{g}} = \frac{Q_2}{10\text{g}} \] where \(Q_1\) is the energy required to heat 5g of aluminum from \(25^{\circ} \mathrm{C}\) to the final higher temperature (in this case, 4.5 J), and \(Q_2\) is the energy required to heat 10g of aluminum over the same temperature interval. Now, we can fill in the values for \(Q_1\), \(m_1\), and \(m_2\): \[ \frac{4.5\text{ J}}{5\text{g}} = \frac{Q_2}{10\text{g}} \]

Solve for Q2

Now, we will solve for \(Q_2\) by cross-multiplying: \[ 4.5\text{ J} \times 10\text{g} = Q_2 \times 5\text{g} \] Divide by 5g on both sides to isolate \(Q_2\): \[ Q_2 = \frac{4.5\text{ J} \times 10\text{g}}{5\text{g}} \]

Calculate the final answer

Now, we just need to calculate the final value for \(Q_2\): \[ Q_2 = \frac{4.5\text{ J} \times 10\text{g}}{5\text{g}} = 9\text{ J} \] So, it will take 9 Joules to warm 10g of aluminum over the same temperature interval.

Key concepts

Energy Calculation

Energy calculation in the context of heat transfer involves determining the amount of energy, usually in joules, needed to change the temperature of a substance. This is often done using the formula for heat transfer: \[ Q = mc\Delta T \]where:

• \( Q \) is the heat energy (in joules) that is added or removed
• \( m \) is the mass of the substance (in grams or kilograms)
• \( c \) is the specific heat capacity of the substance
• \( \Delta T \) is the change in temperature

A key part of energy calculation is understanding these variables and how they impact the amount of energy needed. For example, in our problem, we calculate how much energy is needed to increase the temperature of aluminum.
By understanding that the proportion of the energy required is directly related to the mass, we can easily scale the energy needs for the same temperature change by doubling, halving, or proportionally adjusting the mass.

Specific Heat Capacity

The specific heat capacity of a substance is a measure of how much heat energy it takes to raise the temperature of 1 gram of the substance by 1 degree Celsius. Different materials have different specific heat capacities due to their unique physical and chemical properties. For example, aluminum has a specific heat capacity that allows it to gain and lose heat relatively quickly compared to materials with a higher specific heat capacity.

In mathematical terms, when calculating the energy needed to heat a substance, specific heat capacity (\( c \)) is crucial:

• If the specific heat capacity is high, more energy is needed to produce the same change in temperature.
• Conversely, a lower specific heat would mean less energy is required for the same temperature change.

This concept is pivotal in many practical applications, such as designing cooking utensils or engines, where heat transfer efficiency is key.

Mass and Heat Relationship

Understanding the relationship between mass and heat is vital in thermodynamics. When you heat a substance, the amount of heat required not only depends on the temperature change you desire but also on the mass of the substance being heated.For instance, if you have two different masses of the same material, say aluminum, like in our problem:

• If you want to heat a larger mass, you need proportionally more energy if the temperature change remains the same.
• This can be understood from the formula \( Q = mc\Delta T \) – increasing the mass \( m \) necessitates an increase in \( Q \), the energy, for the same temperature change \( \Delta T \).

In our exercise, we see this relationship practically applied. Doubling the mass from 5g to 10g requires twice the amount of energy to achieve the same temperature change, which simplifies energy calculations by understanding proportional relationships without needing to know the specific heat capacity or exact temperatures.