Problem 45 - 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 \(\mathrm{kg}\) 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 of the metal? (b) Which is more useful for storing thermal energy: 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 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) Specific heat of metal: 214.36 J/kg°C. (b) Water is more useful for thermal storage. (c) The calculated specific heat would be too low.

Step 1: Understanding Heat Transfer

Identify that the heat lost by the metal is equal to the heat gained by the water. This is based on the law of conservation of energy, which states that the energy lost by one body must be gained by another when no external work is done.

Step 2: Write the Heat Transfer Equation

Set up the equation based on the principle: \[ m_m imes c_m imes (T_i - T_f) = m_w imes c_w imes (T_f - T_i) \]Here, \(m_m\) is the mass of the metal (500.0 g or 0.500 kg), \(c_m\) is the specific heat of the metal (which we need to find), \(T_i\) is the initial temperature of the metal (boiling water, approx. 100.0°C), \(T_f\) is the final temperature of both metal and water (22.0°C), \(m_w\) is the mass of water (1.00 kg) and \(c_w\) is the specific heat capacity of water (4.18 J/g°C or 4180 J/kg°C).

Step 3: Substitute Given Values

Substitute the known values into the heat transfer equation: \[ 0.500 imes c_m imes (100.0 - 22.0) = 1.00 imes 4180 imes (22.0 - 20.0) \]This simplifies to: \[ 0.500 imes c_m imes 78 = 8360 \]

Step 4: Solve for Specific Heat of the Metal

Solve for \(c_m\):\[ c_m = \frac{8360}{0.500 imes 78} \]\[ c_m = \frac{8360}{39} \]\[ c_m = 214.36 ext{ J/kg°C} \]

Step 5: Comparing Thermal Energy Storage Potential

Water has a specific heat capacity of 4180 J/kg°C, which is significantly higher than 214.36 J/kg°C for the metal. Hence, water is more useful for storing thermal energy due to its higher specific heat capacity.

Step 6: Impact of Non-Negligible Styrofoam Heat Absorption

If the Styrofoam absorbs some heat, less heat is available to the water from the metal. This means the specific heat of the metal calculated is actually too low, as it appears that the metal lost less heat than it did.

Key Concepts

Heat Transfer

Understanding how heat transfer works is key in thermodynamics. In the context of this exercise, heat transfer refers to the process where the unknown metal transfers its heat to the water, resulting in a temperature change. When two objects of different temperatures come into contact, heat flows from the hotter object to the cooler one until thermal equilibrium is reached. This natural process ensures that energy is conserved within a closed system.

The formula used to calculate this energy exchange is based on the principle of heat transfer:

• The metal loses heat in the process: \[ q_{metal} = m_m \times c_m \times (T_i - T_f) \]*Here, \(q_{metal}\) is the heat lost, \(m_m\) is the mass, \(c_m\) is the specific heat capacity, and \(T_i, T_f\) are the initial and final temperatures.*


• Simultaneously, the water gains the same amount of heat:\[ q_{water} = m_w \times c_w \times (T_f - T_i) \]*This equation mirrors the one used for the metal, with the variables adjusting to describe the water's specifics.*

In essence, the heat lost by the metal equals the heat gained by the water, adhering to the conservation of energy.

Thermal Energy Storage

When it comes to storing thermal energy, specific heat capacity plays a pivotal role. This capacity indicates how much energy a substance can store per unit of mass per degree change in temperature.

In our scenario, we observe that water has a specific heat capacity of 4180 J/kg°C, far exceeding the calculated 214.36 J/kg°C of the metal. This means for every kilogram of these substances heated or cooled by one degree Celsius, water requires significantly more energy compared to the metal.

What does this imply for storage of thermal energy?

• Water is incredibly effective for storing heat due to its higher energy absorption capacity.


• Materials with high specific heat capacities, like water, can absorb and retain heat efficiently, making them ideal for thermal energy storage solutions.

Thus, given a choice between the metal and water for thermal storage, water would be preferred due to its superior capacity to hold energy.

Conservation of Energy

The law of conservation of energy is a fundamental concept in physics, asserting that energy cannot be created or destroyed in an isolated system. When examining systems like the one in the exercise, this principle is at work. Here, energy conservation implies that the total energy lost by the metal must equal the total energy gained by the water.

Practically speaking:

• Initial energy in the metal (due to its high temperature) is converted into energy that raises the water's temperature.
  
• The complete transfer happens within the system confines, with no energy loss to the surrounding environment (assuming a perfectly insulated container, like the Styrofoam).

However, the situation changes if the Styrofoam container doesn't perfectly insulate, absorbing some heat itself. This absorption would mean less energy is available to warm the water, which causes discrepancies in calculated values, like the specific heat capacity of the metal, demonstrating the necessity of considering all energy exchanges in practical applications of energy conservation.