Question

A \(5.00-\mathrm{g}\) sample of aluminum pellets and a \(10.00-\mathrm{g}\) sample of iron pellets are placed together in a dry test tube, and the test tube is heated in a boiling water bath to \(100 .^{\circ} \mathrm{C}\). The mixture of hot iron and aluminum is then poured into \(97.3 \mathrm{g}\) of water at \(22.5^{\circ} \mathrm{C}\). To what final temperature is the water heated by the metals?

Short answer

The final temperature of the water after the hot aluminum and iron pellets are added is approximately \(21.7^{\circ}\mathrm{C}\).

Step-by-step solution

Identify the given information

We are given the following information: - Mass of aluminum (m1): 5.0 g - Mass of iron (m2): 10.0 g - Initial temperature of both metals (T1): 100°C - Mass of water (m3): 97.3 g - Initial temperature of the water (T2): 22.5°C We need to find the final temperature (Tf) of the water.

Write the formula

We'll use the formula for heat transfer, taking into account the specific heat capacities of aluminum, iron, and water: q = mcΔT Where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. The specific heat capacities are: - Specific heat capacity of aluminum (c1): 0.897 J/g°C - Specific heat capacity of iron (c2): 0.449 J/g°C - Specific heat capacity of water (c3): 4.18 J/g°C According to the conservation of energy, q1 + q2 = -q3 (heat lost by metals equals the heat gained by water).

Setup the equation system

Based on the formula, we have: - Heat transfer for aluminum: q1 = m1 × c1 × (Tf - T1) - Heat transfer for iron: q2 = m2 × c2 × (Tf - T1) - Heat transfer for water: q3 = m3 × c3 × (Tf - T2) Now, substitute the given values and the conservation of energy condition: 5.0 × 0.897 × (Tf - 100) + 10.0 × 0.449 × (Tf - 100) = - 97.3 × 4.18 × (Tf - 22.5)

Solve the equation for the final temperature

Simplify the equation and solve for Tf: 4.485(Tf - 100) + 4.49(Tf - 100) = - 406.834(Tf - 22.5) 8.975Tf - 897.5 = - 406.834Tf + 9140.265 Combine like terms and solve for Tf: 415.809Tf = 9037.765 Tf ≈ 21.7°C

Interpret the result

The final temperature of the water after the hot aluminum and iron pellets are added is approximately 21.7°C. This is a bit lower than the initial temperature of water because the water is having more heat capacity and the mass ratio between metal and water also played a significant role in determining the final temperature.

Key concepts

Understanding Specific Heat Capacity

Specific heat capacity is an important property that helps us understand how substances absorb and transfer heat. It refers to how much energy is needed to raise the temperature of a unit mass of a substance by 1°C.
It's an indication of how resistant a substance is to temperature changes.

• Specific heat capacity is denoted by the symbol \( c \).
• Each substance has its own specific heat capacity based on its physical structure and composition.
• In our exercise, aluminum has a specific heat capacity of 0.897 J/g°C, whereas iron has one of 0.449 J/g°C and water has a higher specific heat capacity of 4.18 J/g°C.

A higher specific heat capacity means a substance requires more energy to change its temperature. That’s why water, with its high specific heat capacity, acts as a great temperature buffer.

Conservation of Energy in Heat Exchange

The principle of conservation of energy is a huge focus in the study of heat transfer. This principle states that energy cannot be created or destroyed, only transformed from one form into another.
In the context of heat transfer, it implies that the total heat lost by the hot substances must equal the total heat gained by the colder substances.

• When the aluminum and iron pellets are added to the water, the heat they lose is gained by the water.
• The equation we use is \( q_1 + q_2 = -q_3 \), ensuring that the energy leaving the metals equals the energy entering the water.
• This balance allows us to calculate the final temperature by equating energy loss of the metals with the energy gained by water.

Understanding this energy balance is crucial for solving problems involving multiple substances with varying temperatures.

The Role of Temperature Change

Temperature change is a central aspect of heat transfer problems, as it is the result we often look to find.
In our given exercise, temperature change is represented as \( \Delta T \), which translates to "change in temperature."

• The formula \( \Delta T = T_f - T_i \) helps us quantify how much a substance's temperature has shifted.
• Here, \( T_f \) represents the final temperature, and \( T_i \) is the initial temperature of the substance.
• The temperature change experienced by aluminum and iron results in their heat loss, while water’s temperature change indicates the heat gained.

Discovering the final temperature \( T_f \) involves solving an equation where each metal's temperature change relates to how much energy they contribute to the water.