Question

Some solar-heated homes use large beds of rocks to store heat. (a) How much heat is absorbed by \(100.0 \mathrm{~kg}\) of rocks if their temperature increases by \(12^{\circ} \mathrm{C} ?\) (Assume that \(c=0.82 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) (b) Assume that the rock pile has total surface area \(2 \mathrm{~m}^{2}\). At maximum intensity near the earth's surface, solar power is about 170 watts \(/ \mathrm{m}^{2}\). (1 watt = \(1 \mathrm{~J} / \mathrm{s}\).) How many minutes will it take for solar power to produce the \(12^{\circ} \mathrm{C}\) increase in part (a)?

Short answer

Answer: The heat absorbed by the rocks is 984000 J, and it takes approximately 48.24 minutes for solar power to produce the temperature increase.

Step-by-step solution

Calculate the amount of heat absorbed

To calculate the amount of heat absorbed by the rocks, we will use the formula Q = mcΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Given the information: - mass (m) = 100 kg - specific heat capacity (c) = 0.82 J/g°C (we need to convert it to J/kg°C) - change in temperature (ΔT) = 12°C First, let's convert the specific heat capacity from J/g°C to J/kg°C: 1 kg = 1000 g, so \(c_{kg} = 0.82 \frac{\mathrm{J}}{\mathrm{g}\cdot \mathrm{C}} \cdot \frac{1000 \mathrm{g}}{1\mathrm{kg}} = 820 \frac{\mathrm{J}}{\mathrm{kg}\cdot \mathrm{C}}\) Now, let's calculate the heat (Q): \(Q = mcΔT = (100 \mathrm{~kg}) \cdot (820 \frac{\mathrm{J}}{\mathrm{kg}\cdot \mathrm{C}}) \cdot (12 \mathrm{~C}) = 984000 \mathrm{~J}\)

Determine the time it takes for solar power to produce the temperature increase

Now that we have calculated the amount of heat absorbed (984000 J), we can use the solar power information to determine how long it takes for the solar power to produce this heat. We're given the following information: - rock pile's total surface area (A) = 2 m² - maximum solar power intensity near Earth's surface (P) = 170 W/m² First, let's find the total solar power (P_total) received by the rock pile: \(P_{total} = PA = (170 \frac{\mathrm{J}}{\mathrm{s}\cdot \mathrm{m}^{2}}) \cdot (2 \mathrm{~m}^{2}) = 340\ \mathrm{W}\) We know that power (P) is equal to the energy (E) divided by time (t): \(P = \frac{E}{t}\) We want to find the time (t) it takes for the solar power to produce the energy (Q) calculated in step 1, so we can rearrange the formula: \(t = \frac{E}{P}\) Now let's plug in the values and find the time (t) in seconds: \(t = \frac{984000\ \mathrm{J}}{340\ \mathrm{W}} = 2894.12\ \mathrm{s}\) Finally, let's convert our result to minutes: \(\mathrm{Time} = \frac{2894.12\ \mathrm{s}}{60\ \mathrm{s/min}} = 48.24\ \mathrm{minutes}\) In conclusion: (a) The heat absorbed by 100 kg of rocks that experience a temperature increase of 12°C is 984000 J. (b) It takes approximately 48.24 minutes for solar power to produce the 12°C temperature increase.

Key concepts

Specific Heat Capacity

Specific heat capacity is a fundamental concept in heat transfer and thermodynamics. It refers to the amount of heat energy required to raise the temperature of one kilogram of a substance by one degree Celsius. Different materials have different specific heat capacities, which affect how they absorb or release heat.
To calculate the heat absorbed or released by a substance, we use the formula:

• \( Q = mc\Delta T \)

In this formula, \( Q \) represents the heat energy (in joules), \( m \) is the mass (in kilograms), \( c \) is the specific heat capacity (in J/kg°C), and \( \Delta T \) is the change in temperature (in degrees Celsius). Understanding this concept helps in analyzing how different materials, like rocks, will respond to changes in temperature.

Solar Power

Solar power harnesses the energy from sunlight to generate electricity or heat, making it a key player in renewable energy. The intensity of solar power depends on location, time of day, and atmospheric conditions.
Solar power near the Earth's surface can reach levels of about 170 W/m² during peak sunlight. This measurement represents the amount of energy received per second per square meter. When using solar power for heating, like in the rock bed example, it's important to consider the surface area that receives sunlight.
For instance, if the rock pile has a surface area of 2 m², the total solar power received is calculated as:

• \( P_{\text{total}} = P \times A \)
• Where \( P \) is the solar power intensity and \( A \) is the surface area.

Energy Calculation

Calculating energy in the context of heat transfer often involves determining how much energy (in joules) is needed to change the temperature of a substance. This is often connected to how materials respond when absorbing or releasing energy.
The formula to calculate energy required is crucial: \( Q = mc\Delta T \). This formula tells us that energy depends on mass, specific heat capacity, and temperature change.
For solar heating applications, once the amount of energy needed is known, the time required to absorb this energy from sunlight is determined by:

• \( t = \frac{E}{P} \)

Here, \( E \) is the energy required (the result from the heat equation), and \( t \) is the time in seconds. Knowing this helps in determining how efficient solar power systems can be.

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat and other forms of energy. It is crucial for understanding how energy transformations occur, especially in systems using solar power for heating or other energy transfer methods.
In solar power systems, thermodynamics principles help explain how energy from the sun is stored and used. For example, in the context of heating rocks, energy is absorbed and stored as internal energy, raising the temperature of the rocks.
Overall, principles of thermodynamics aid in designing systems that maximize energy efficiency, reduce loss, and effectively use resources like solar energy. Focus on the law of conservation of energy underscores that energy can neither be created nor destroyed, only transformed, which is vital when considering renewable energy solutions.