Problem 43 - College Physics

Question

\(\bullet\) A 0.500 kg chunk of an unknown metal that has been in boiling water for several minutes is quickly dropped into an insulating Styrofoam TM beaker containing 1.00 kg of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\) (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 capacity of the metal? (b) Which is more useful for storing energy from heat, this metal or an equal weight of water? Explain. (c) What if the heat absorbed by the Styrofoam"" actually is not negligible. How would the specific heat capacity you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain your reasoning.

Step-by-Step Solution

Answer

The specific heat capacity of the metal is approximately 0.214 J/g°C. Water is better at storing heat than the metal. If Styrofoam absorbed heat, the metal's heat capacity was underestimated.

Step 1: Set Up the Heat Balance Equation

In the problem, the metal releases heat while the water absorbs it. Since the system is insulated, the heat lost by the metal is equal to the heat gained by the water. Using the heat transfer formula \(q = mc\Delta T\), we can set up the equation: \[m_{\text{metal}}c_{\text{metal}}(T_f - T_i)_{\text{metal}} = m_{\text{water}}c_{\text{water}}(T_f - T_i)_{\text{water}}\].

Step 2: Identify Known Values

We know \(m_{\text{metal}} = 0.500\, \text{kg}\), \(m_{\text{water}} = 1.00\, \text{kg}\), \(c_{\text{water}} = 4.18\, \text{J/g}\cdot\text{°C}\), initial temperature of the water \(T_i = 20.0^\circ C\), and final temperature of the water \(T_f = 22.0^\circ C\). The initial temperature of the metal is \(100.0^\circ C\) (boiling point of water).

Step 3: Solve for the Specific Heat Capacity of the Metal

Substitute the known values into the equation: \(0.500\,c_{\text{metal}}(22.0 - 100.0) = 1.00\, \times 4.18 \times (22.0 - 20.0)\). Simplifying gives: \(-39.0\, \times c_{\text{metal}} = 8.36\). Solving for \(c_{\text{metal}}\) gives \(c_{\text{metal}} \approx 0.214\, \text{J/g}\cdot\text{°C}\).

Step 4: Compare Energy Storage for Metal and Water

Water has a specific heat capacity of \(4.18\, \text{J/g}\cdot\text{°C}\), which is much larger than the specific heat capacity of the metal \(0.214\, \text{J/g}\cdot\text{°C}\). Hence, water is more useful for storing energy from heat compared to the metal.

Step 5: Analyze the Effect of Neglecting Heat Absorbed by the Styrofoam

If the Styrofoam absorbs some heat, then water absorbs less heat than calculated. Thus, the calculated specific heat capacity of the metal would be too small. This is because less heat transfer to water implies less heat was released by the metal.

Key Concepts

Heat Transfer Equation

In exercises involving the exchange of heat between substances, the heat transfer equation is crucial for analyzing the changes in temperature. This equation is formulated as: \( q = mc\Delta T \). It describes how heat \( q \) is transferred based on mass \( m \), specific heat capacity \( c \), and the temperature change \( \Delta T \). The pivotal concept here is energy conservation, which means that in an isolated system, energy lost by one body must be gained by another.

In our current scenario involving an unknown metal and water, the energy lost by the metal as it cools down is equal to the energy gained by the water heating up. We set up the equation: - \( m_{\text{metal}}c_{\text{metal}}(T_f - T_i)_{\text{metal}} = m_{\text{water}}c_{\text{water}}(T_f - T_i)_{\text{water}} \).

This equation allows us to solve for the metal's specific heat capacity \( c_{\text{metal}} \) because the rest of the variables are given. It emphasizes the importance of these variables in quantifying and understanding energy exchange.

Energy Storage Comparison

In the context of specific heat capacity, it is critical to understand how different materials store energy. Specific heat capacity is defined as the amount of heat required to change the temperature of a unit mass by one degree Celsius.

Given the values in the problem, water has a specific heat capacity of \(4.18\, \text{J/g} \cdot \text{°C} \), while the calculated value for the unknown metal is significantly lower at \( 0.214\, \text{J/g} \cdot \text{°C} \).

What does this mean practically?

• Water stores more energy than the same mass of metal when both are subjected to the same temperature change.
• This property makes water an excellent substance for thermal energy storage, with applications that include heating and cooling systems.

The comparison highlights the necessity to choose the right material based on energy storage requirements in practical applications.

Error Analysis

Error analysis is essential when making precise measurements and calculations. In the given problem, we initially assumed that the Styrofoam absorbs a negligible amount of heat, but what if it doesn't?

If the Styrofoam actually absorbs a measurable amount of heat, it would affect the system's total heat balance. Here’s how this changes the scenario:

• The water would absorb less heat than initially calculated, as some energy would be absorbed by the Styrofoam.
• This would make the calculated specific heat capacity of the metal too small, because less heat gain by water implies less heat loss by the metal than assumed.

Understanding the impact of this potential error helps in improving future experimental setups by ensuring that all possible heat transfers are considered.

Thermal Equilibrium

Reaching thermal equilibrium is a fundamental concept in thermodynamics that occurs when two substances in physical contact reach the same temperature and no further heat flows between them.

In the current exercise, thermal equilibrium was reached when the metal transferred heat to the water until their temperature stabilized at \(22.0^{\circ} \text{C} \). This process illustrates how energy transfer continues until both substances are at the same temperature.

• Thermal equilibrium is important in real-world applications, such as designing thermal systems where maintaining even temperatures is crucial.
• The time taken to reach equilibrium can also provide insights into the thermal properties of the materials involved.

This concept not only aids in understanding the exchange and balance of energy but also helps in predicting how different materials will behave under heat transfer conditions.