Aluminum 0.897 Soil 0.80 Granite 0.790 Iron 0.450 Copper 0.385 Silver 0.233 Lead 0.129 A. 4.485 J B. 2691 J C. $13,455 J$ 0.13455 .1
Answer (2)
Assume the problem asks for heat calculation using Q = m c Δ T .
Analyze option A: Q = 5 × 0.897 × 1 = 4.485 J (Aluminum, 5g, 1°C).
Analyze options B and C, noting they require larger mass or temperature changes.
Conclude that option A is the most plausible based on the given data: 4.485 J .
Explanation
Understanding the Problem We are given a table of specific heat capacities for various materials. We are also given three options: A. 4.485 J, B. 2691 J, C. 13 , 455 J . The problem is incomplete, as it does not specify what we need to calculate. To make an educated guess, let's assume we are asked to calculate the heat required to raise the temperature of a certain mass of a material by a certain temperature difference. The formula for heat transfer is Q = m c Δ T , where Q is the heat transfer, m is the mass, c is the specific heat capacity, and Δ T is the change in temperature.
Analyzing Option A Let's analyze option A, 4.485 J. Looking at the table, Aluminum has a specific heat capacity of 0.897 J/g°C. If we assume a mass of 5g and a temperature change of 1°C, the heat required would be: Q = m c Δ T = 5 × 0.897 × 1 = 4.485 J This matches option A.
Analyzing Option B Let's analyze option B, 2691 J. It's difficult to find a combination of mass, specific heat, and temperature change that results in this value without more context. However, we can try to find some reasonable values. For example, if we consider Iron (c = 0.450 J/g°C), we would need m Δ T = 0.450 2691 = 5980 . This could be a mass of 598g and a temperature change of 10°C, or other combinations.
Analyzing Option C Let's analyze option C, 13 , 455 J . Again, it's difficult to find a direct combination. If we consider Lead (c = 0.129 J/g°C), we would need m Δ T = 0.129 13455 ≈ 104302 . This would require a very large mass or temperature change.
Conclusion Based on the given options and the specific heat capacities, option A (4.485 J) seems the most plausible if we assume a mass of 5g of Aluminum and a temperature change of 1°C. Without more information, we can only make an educated guess.
Examples
Understanding specific heat capacity is crucial in many real-world applications. For instance, when designing cookware, engineers consider the specific heat capacity of the material to ensure even heat distribution and prevent burning. Similarly, in construction, the thermal properties of materials like concrete and steel are carefully chosen to regulate building temperatures and reduce energy consumption. By understanding how different materials respond to heat, we can create more efficient and comfortable environments.
The calculated heat required is 4.485 J, which corresponds to option A. This was derived using Aluminum's specific heat capacity with a mass of 5g and a temperature change of 1°C. The other options require either unreasonably large masses or temperature changes to be feasible.
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