Problem 35 - University Physics with Modern Physics

Question

A 500.0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at room temperature (20.0\(^\circ\)C). After waiting and gently stirring for 5.00 minutes, you observe that the water’s temperature has reached a constant value of 22.0\(^\circ\)C. (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

Step-by-Step Solution

Answer

(a) 0.214 J/g°C. (b) Water; it has higher specific heat. (c) Too large; some heat went to the Styrofoam.

Step 1: Understanding the Heat Transfer Concept

We are dealing with a heat transfer problem where heat lost by the metal equals heat gained by water, provided no heat is lost to the environment. This is based on the principle of conservation of energy.

Step 2: Setting Up the Equation

For the metal and the water, the heat transfer can be described by the equation: \[ q_{metal} = -q_{water} \]Where \( q \) represents the heat exchanged. Specifically, \[ m_{metal} \cdot c_{metal} \cdot (T_{final} - T_{initial, metal}) = - m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water}) \]with:- \( m_{metal} = 500.0 \text{ g} \)- \( T_{initial, metal} = 100.0^\circ C \) (boiling water temperature)- \( T_{final} = 22.0^\circ C \)- \( m_{water} = 1000.0 \text{ g} \)- \( T_{initial, water} = 20.0^\circ C \)- \( c_{water} = 4.18 \frac{J}{g^\circ C} \).

Step 3: Solving for Specific Heat of the Metal

Rearranging the equation to solve for the metal’s specific heat: \[ c_{metal} = \frac{- m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water})}{m_{metal} \cdot (T_{final} - T_{initial, metal})} \].Substituting known values:\[ c_{metal} = \frac{- (1000 \text{ g}) \times 4.18 \frac{J}{g^\circ C} \times (22.0^\circ C - 20.0^\circ C)}{500.0 \text{ g} \times (22.0^\circ C - 100.0^\circ C)} \].Calculating:\[ c_{metal} = \frac{(1000) \times 4.18 \times 2}{500 \times (-78)} = 0.214 \frac{J}{g^\circ C} \].

Step 4: Comparing Specific Heats for Thermal Energy Storage

The specific heat of water (4.18 \(J/g^\circ C\)) is significantly higher than for the metal (0.214 \(J/g^\circ C\)). Thus, for an equal weight, water can store more thermal energy than the metal.

Step 5: Effect of Non-negligible Styrofoam Heat Absorption

If the Styrofoam absorbed some heat, the water would gain less heat. Thus, the calculated specific heat of the metal would be too high, because we assumed all heat went to the water.

Key Concepts

Heat Transfer

Heat transfer is a fundamental principle in physics, describing the movement of thermal energy from one body or material to another. Usually, thermal energy moves from a hotter object to a colder one until thermal equilibrium is achieved. In the problem at hand, heat transfer occurs between a heated metal and cool water. When the metal object, initially at a higher temperature, is placed into the water, it loses heat energy while the water absorbs this energy, slightly raising its temperature.
To understand the heat transfer in this scenario, the principle of energy conservation is applied. The thermal energy lost by the metal is equal to the thermal energy gained by the water, as long as we assume no heat is lost to the surroundings or absorbed by the Styrofoam beaker. This idea can be summarized in the following equation:
\( q_{metal} = -q_{water} \), where \( q \) represents the exchanged heat in joules. By applying this concept, we can calculate critical values such as the specific heat capacity of an unknown metal when it's submerged in water.

Thermal Energy Storage

Thermal energy storage refers to the material's ability to retain heat. This ability is quantified by specific heat capacity, which indicates how much heat energy a material can absorb per unit mass before its temperature increases. The higher the specific heat capacity, the more energy the material can store.
In our study, water acts as a better storage medium for thermal energy when compared with the unknown metal, due to its specific heat capacity being significantly greater (4.18 \(J/g^\circ C\) for water versus 0.214 \(J/g^\circ C\) for the metal).
This implies that for an equal mass, water can absorb and retain more heat energy, making it more efficient for thermal storage.

• For example, in terms of everyday applications, water heats and cools more slowly than metals.
• Thus, water is frequently used in systems designed for regulating temperature and storing thermal energy, such as in heating systems and cooling processes.

The comparison helps in choosing materials for different applications based on their thermal energy storage capabilities.

Conservation of Energy

Conservation of energy is a vital principle stating that energy cannot be created or destroyed in an isolated system, only transformed. This concept ensures that the total energy in a closed system remains constant.
When applying this to a thermal context, particularly our exercise, it states that the amount of heat lost by the metal will equal the heat gained by the water when no external heat is added or lost. The fundamental equation: \( m_{metal} \cdot c_{metal} \( T_{final} - T_{initial, metal} \) = - m_{water} \cdot c_{water} \( T_{final} - T_{initial, water} \) \), reflects this balance.
Recognizing this principle is key in conducting accurate thermal calculations in physical scenarios.

• It is the basis upon which many engineering applications are designed, ensuring systems are efficient and sustainable by accounting for all heat transfers.
• In our daily lives, it helps improve energy systems, like refrigerators, turbines, and combustion engines, making sure they follow conservation laws to optimize performance.

With this foundational knowledge, you can properly analyze circumstances involving energy exchanges, such as identifying potential calculation errors caused by overlooking heat absorption by the beaker.