Question

Two solid objects, \(\mathrm{A}\) and \(\mathrm{B}\), are placed in boiling water and allowed to come to temperature there. Each is then lifted out and placed in separate beakers containing \(1000 \mathrm{~g}\) water at \(10.0^{\circ} \mathrm{C}\). Object \(\mathrm{A}\) increases the water temperature by \(3.50^{\circ} \mathrm{C} ; \mathrm{B}\) increases the water temperature by \(2.60{ }^{\circ} \mathrm{C}\). (a) Which object has the larger heat capacity? (b) What can you say about the specific heats of \(\mathrm{A}\) and \(\mathrm{B}\) ?

Short answer

(a) Object A has a larger effect on the water temperature compared to B, which means it has a smaller heat capacity. This is because a smaller heat capacity indicates that a smaller amount of heat is needed to produce the same change in temperature. (b) Since object A has a smaller heat capacity, it should also have a smaller specific heat. The specific heat is directly proportional to the heat capacity and describes the amount of heat required to raise the temperature of an object by 1°C per unit mass.

Step-by-step solution

Understanding the heat transfer equation

The heat transfer equation states that the heat gained by one substance is equal to the heat lost by another substance. In this case, the heat gained by the water is equal to the heat lost by the solid objects A and B. Mathematically, the equation looks like this: \(q_\textrm{gained} = q_\textrm{lost}\) In this exercise, we will be using the concept of heat capacity (C) and specific heat (c) to compute the heat transfer.

Determine the heat gained by water

First, let's determine the heat gained by the water (q_1 and q_2) in the two beakers. The formula for heat gained is given by: \(q_\textrm{gained} = mc\Delta T\) Where m is the mass of the water, c is the specific heat capacity of water, and ∆T is the change in temperature. We are given the mass of water in each beaker (1000 g) and ∆T based on the temperature increase (3.50°C for A and 2.60°C for B). We also know that the specific heat capacity of water is approximately 4.18 J/g°C. Using these values, we can now calculate the heat gained by the water in each beaker: For A (q_1): \(q_1 = (1000\;\textrm{g})(4.18\;\displaystyle\frac{\textrm{J}}{\textrm{g}^\circ\textrm{C}})(3.50^\circ\;\textrm{C})\) For B (q_2): \(q_2 = (1000\;\textrm{g})(4.18\;\displaystyle\frac{\textrm{J}}{\textrm{g}^\circ\textrm{C}})(2.60^\circ\;\textrm{C})\)

Calculating the heat lost by A and B

Since the heat gained by the water is equal to the heat lost by the solid objects A and B, we know that: \(q_1 = q_\textrm{Alost}\) \(q_2 = q_\textrm{Blost}\) Now, we will find the heat capacity (C) of solid objects A and B by calculating the heat lost using the formula: \(q_\textrm{lost} = C\Delta T\)

Comparing the heat capacities and specific heats of A and B

(a) To find which object has a larger heat capacity, we must compare the heat capacities of objects A and B. Since A had a larger effect on the water temperature compared to B, we can infer that object A has a smaller heat capacity. This is because a smaller heat capacity means that a smaller amount of heat is needed to produce the same change in temperature. (b) To make a statement about the specific heat of objects A and B, we cannot give an exact value. However, we know that since object A has a smaller heat capacity, it should also have a smaller specific heat. This is because the specific heat is directly proportional to the heat capacity. The specific heat of an object describes the amount of heat required to raise the temperature of that object by 1°C per unit mass. Since A has a smaller heat capacity and affects the temperature of the water more significantly than B, it should also have a smaller specific heat.

Key concepts

Specific Heat

Specific heat capacity is an essential concept in understanding how various materials respond to heat. It is the amount of heat required to raise the temperature of one gram of a substance by 1°C. Different materials have varying abilities to hold and transfer heat, which is represented by this characteristic.
The formula for specific heat is:\[c = \frac{q}{m \Delta T}\]Where:- \( c \) is the specific heat capacity,- \( q \) is the heat added,- \( m \) is the mass of the substance,- \( \Delta T \) is the change in temperature.

• Higher specific heat implies a substance requires more energy to change its temperature.
• Lower specific heat means a substance changes temperature more quickly with the same amount of energy.

For example, in the exercise, object A affected the water temperature more significantly than B, indicating that A might have a lower specific heat than B. Thus, while A absorbs or loses heat, its temperature changes more readily, signifying a smaller specific heat.

Heat Transfer

Heat transfer is the flow of thermal energy from one object to another due to a temperature difference. This concept is vital in thermodynamics and daily applications when describing how energy moves between systems.
Heat transfer can occur in three ways:- Conduction is the transfer through a solid material from high-energy particles to lower-energy neighbors.- Convection involves the movement of heat through fluids, like air or water, by the motion of the fluid itself.- Radiation transfers heat through electromagnetic waves and can occur across a vacuum.
In our problem, heat transfer happens through the exchange of thermal energy between hot objects (A and B) and cooler water. The heat they lose (the objects), causes the water temperature to rise, demonstrating this critical energy balance. The equations we use, such as the heat gained or lost \( q = mc\Delta T \), help quantify and predict these changes.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy in systems. Its principles explain energy flow in processes and the laws governing such conversions.

• First Law (Conservation of Energy): Energy cannot be created or destroyed, only transformed. This is evident when the heat lost by objects A and B is gained by the water, keeping total energy constant.
• Second Law: Energy transfer processes favor increasing entropy, meaning systems naturally progress towards thermal equilibrium, where temperatures equalize.

In practical exercises, such as this, thermodynamic principles can predict how energy will transfer. The heat capacities provided insights into intrinsic material properties, reflecting how each material stores and distributes energy under thermal processes.