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)?

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): \(Q = mcΔT = (100 \mathrm{~kg}) \cdot (820 \frac{\mathrm{J}}{\mathrm{kg}\cdot \mathrm{C}}) \cdot (12 \mathrm{~C}) = 984000 \mathrm{~J}\)

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.