Question

If it takes \(103 \mathrm{J}\) of energy to warm a certain mass of iron from \(25^{\circ} \mathrm{C}\) to \(50 .^{\circ} \mathrm{C},\) then it will take ____ J to warm the same mass of iron from \(25^{\circ} \mathrm{C}\) to \(75^{\circ} \mathrm{C}\).

Short answer

It will take \(206 \mathrm{J}\) of energy to warm the same mass of iron from \(25^{\circ} \mathrm{C}\) to \(75^{\circ}\mathrm{C}\).

Step-by-step solution

Identify the known variables and formula

We know the following variables: - The initial energy E1 = 103 J - The initial change in temperature ΔT1 = 50°C - 25°C = 25°C - The final change in temperature ΔT2 = 75°C - 25°C = 50°C We will use the formula E = mcΔT.

Calculate the ratio of ΔT2 to ΔT1

Since we're given the energy required for the initial change in temperature and want to find the energy required for the final change in temperature, we can find the ratio of ΔT2 to ΔT1: \( \frac{\Delta T_2}{\Delta T_1} = \frac{50^\circ \mathrm{C}}{25^\circ \mathrm{C}} = 2 \)

Find the energy required for ΔT2

Since the mass and specific heat capacity remains constant, the energy required will be directly proportional to the change in temperature. Therefore, we can multiply the initial energy by the ratio of ΔT2 to ΔT1 to find the energy required for ΔT2: E2 = E1 × 2 = 103 J × 2 = 206 J Hence, it will take 206 J of energy to warm the same mass of iron from 25°C to 75°C.

Key concepts

Energy Calculation

When dealing with energy calculations involving heating substances, one often uses the specific heat capacity formula, which is expressed as:\[ E = mc\Delta T \]Here's a breakdown of each component:

• \( E \) is the energy required to change the temperature, measured in joules (J).
• \( m \) is the mass of the substance being heated, typically in kilograms (kg).
• \( c \) is the specific heat capacity, which indicates how much energy is needed to raise the temperature of 1 kg of the substance by 1°C, and is measured in J/(kg°C).
• \( \Delta T \) is the change in temperature, expressed in degrees Celsius (°C).

In our example, the mass and the specific heat capacity of iron are not explicitly given, but they remain unchanged as the temperature increases. Therefore, when the change in temperature doubles (from 25°C to 50°C and then from 25°C to 75°C), the energy needed doubles as well. Thus, if it takes 103 J to increase the temperature by 25°C, it will require 206 J to increase it by 50°C.

Temperature Change

Temperature change is the difference between the final and initial temperatures, represented as \( \Delta T \). In processes like heating, this indicates how much the temperature rises. The formula for change in temperature is:\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]For the given problem, we have:

• The initial change in temperature \( \Delta T_1 \) is from 25°C to 50°C, resulting in a \( \Delta T \) of 25°C.
• The final change in temperature \( \Delta T_2 \) extends from 25°C to 75°C, giving a \( \Delta T \) of 50°C.

Understanding how temperature change relates to energy usage is crucial. As seen, the greater the change in temperature, the more energy is required, assuming mass and specific heat capacity are constant. This is why a doubling in the temperature change results in an energy doubling in our scenario.

Iron Properties

Iron is a common metal used in various applications, and understanding its thermal properties is essential for energy calculations. One key property is its specific heat capacity, typically about 0.449 J/(g°C) or 449 J/(kg°C). This means that raising the temperature of 1 kg of iron by 1°C requires 449 J of energy. Here are some quick facts about iron's properties in thermal contexts:

• Iron's specific heat capacity is relatively low compared to water, meaning it heats up quickly.
• In the given problem, even without directly knowing the mass or specific heat capacity, we can infer behavior through the formula \( E = mc\Delta T \).
• Keeping the mass of iron constant means any energy changes stem from differences in the temperature change, as seen in our scenario.

Understanding these aspects of iron helps in predicting how it will respond to heating and incorporating it into broader energy-related calculations.