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 A and B?

Short answer

Object A has the larger heat capacity because it has a higher value of ΔQ (14630 J) compared to object B (10868 J). We cannot determine the exact specific heats of objects A and B, but we can say that object A has a lower specific heat than object B, assuming both objects have the same mass.

Step-by-step solution

Calculate the heat absorbed by the water in each case

In order to find the heat absorbed by the water in each beaker, use the equation ΔQ = mcΔT, where m is the mass of the water, c is the specific heat of water, and ΔT is the change in temperature. The mass m is given as 1000 g, the specific heat c of water is 4.18 J/g°C, and temperature changes ΔT are given as 3.50°C and 2.60°C for objects A and B, respectively. For object A: ΔQ_A = (1000 g) × (4.18 J/g°C) × (3.50°C) For object B: ΔQ_B = (1000 g) × (4.18 J/g°C) × (2.60°C) Calculate these values: ΔQ_A = 14630 J ΔQ_B = 10868 J

Determine the heat capacity of each object

Since the objects are placed in separate beakers, we can assume that the heat lost by the objects is equal to the heat gained by the water. Therefore, the heat capacity of each object can be expressed as the heat absorbed by the water in each beaker. Heat capacity of object A: C_A = ΔQ_A = 14630 J Heat capacity of object B: C_B = ΔQ_B = 10868 J

Compare the heat capacities

Now we need to compare the heat capacities of objects A and B in order to determine which object has the larger heat capacity. C_A > C_B (14630 J > 10868 J) So, object A has the larger heat capacity.

Determine the specific heats of objects A and B

We cannot calculate the exact specific heats of objects A and B, as the exercise does not provide enough information, such as the mass of the objects. However, we can make a qualitative statement concerning their specific heats. Assuming both objects absorb the same amount of heat from the boiling water, we can state that since object A has a larger heat capacity, it will have lower specific heat than object B. This is due to the relationship between heat capacity (C), mass (m), and specific heat (c): C = mc. If C_A > C_B and m_A = m_B, then c_A < c_B. Therefore, we can conclude that the specific heat of object A is lower than that of object B.

Key concepts

Specific Heat

Specific heat is a fundamental concept in thermodynamics. It refers to the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius. It is usually denoted by the symbol \( c \), and its unit is typically Joules per gram per degree Celsius (\( \, \text{J/g°C} \)).

To understand specific heat, consider water. Water has a high specific heat, which means it can absorb a lot of heat without a significant change in temperature. This property is why water is an excellent medium for regulating temperatures in nature and human-made systems.

In the context of the exercise, the difference in specific heats between objects A and B gives clues about their materials without needing the masses of the objects. If both objects are assumed to have absorbed the same amount of heat from the boiling water, object A, with a larger heat capacity but perhaps higher mass, would have a lower specific heat than object B. This means object A heats up more slowly than object B, suggesting object A could be denser or made from a substance with a naturally lower specific heat.

Calorimetry

Calorimetry is the science of measuring the amount of heat. This process is crucial in experiments and energy calculations where heat exchange is involved, such as the exercise here. A calorimeter is a device used to measure the heat absorbed or released during a physical or chemical process.

In the given exercise, calorimetry principles help determine the heat absorbed by the water when objects A and B are placed in separate beakers. By knowing the mass and specific heat of water, and observing the temperature change, the heat transferred can be calculated using the formula: \[ \Delta Q = mc\Delta T \\]

• \( \Delta Q \) represents the heat absorbed
• \( m \) stands for the mass of the water
• \( c \) is the specific heat of water
• \( \Delta T \) is the temperature change measured

Therefore, the calculations in the exercise relied on principles of calorimetry to establish the differing heat capacities of the two objects via the heat changes in water.

Heat Transfer

Heat transfer is the movement of heat from one body or system to another. It can occur through three main processes: conduction, convection, and radiation. In this exercise, heat transfer occurs primarily through conduction, as the heat directly transfers from the hot objects to the cooler water when they are in contact.

The effectiveness of heat transfer depends on the temperature difference between the two bodies and their respective heat capacities. The formula \( \Delta Q = mc\Delta T \) used in the exercise embodies the principles of heat transfer, showcasing how changes in temperature are related to the specific heat and mass of the water.

In practical terms, understanding heat transfer is essential for solving problems in calorimetry. This is because it helps predict how temperature will change in a substance when heat is added or removed. For example, in our exercise, object A transferred more heat to the water than object B, leading to a more significant temperature increase in the water. This indicates how heat transfer varies depending on the material properties of the objects involved.