The enthalpy of fusion of water is \(6.01 \mathrm{~kJ} / \mathrm{mol}\). Sunlight striking Earth's surface supplies \(168 \mathrm{~W}\) per square meter \((1 \mathrm{~W}=1 \mathrm{watt}=1 \mathrm{~J} / \mathrm{s}) .(\) a) Assuming that melting of ice is due only to energy input from the Sun, calculate how many grams of ice could be melted from a 1.00 square meter patch of ice over a \(12-\mathrm{h}\) day. \((\mathbf{b})\) The specific heat capacity of ice is \(2.032 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}\). If the initial temperature of a 1.00 square meter patch of ice is \(-5.0^{\circ} \mathrm{C},\) what is its final temperature after being in sunlight for \(12 \mathrm{~h}\), assuming no phase changes and assuming that sunlight penetrates uniformly to a depth of \(1.00 \mathrm{~cm} ?\)

First, we need to calculate the total energy input from the Sun in 12 hours. Energy input rate is 168 J/s (given). We need to find the total energy input in 12 hours. 12 hours = 12 * 60 * 60 seconds Energy input = (Energy input rate) x (Time) Energy input = 168 J/s x (12 * 60 * 60 s) \[Energy\:input = 7,257,600\:J\] We are given the enthalpy of fusion of water, denoted by ∆H, which is 6.01 kJ/mol. To convert it to joules, multiply by 1000. ∆H = 6.01 kJ/mol * 1000 J/kJ \[\Delta H = 6,010\:J/mol\] Next, we need to find the number of moles (n) of ice that can be melted using this energy. n = (Energy input) / ∆H n = 7,257,600 J / 6,010 J/mol \[n \approx 1,207\:moles\] Now, we can convert moles to grams using the molar mass of water (18 g/mol). Mass = n * (molar mass of water) Mass = 1,207 moles * 18 g/mol \[Mass \approx 21,726\:grams\] So, about 21,726 grams of ice can be melted from a 1.00 square meter patch of ice over a 12-hour day.

Now, we are given the specific heat capacity of ice (c) which is 2.032 J/g°C. We also know the initial temperature of the ice (T_initial) is -5.0°C. Given the ice is 1.00 square meter in area and the sunlight penetrates uniformly to a depth of 1.00cm, we can calculate the volume of the ice. Volume of ice = Area x Depth \[Volume\:of\:ice = 1.00\:m^2 * 0.01\:m = 0.01\:m^3\] Next, we will assume the density of ice as 917 kg/m³, which is a common value for ice. Now, we can calculate the mass of the ice patch: Mass of ice = (Volume of ice) x (Density of ice) \[Mass\:of\:ice = 0.01\:m^3 * 917\:kg/m^3 = 9.17\:kg\] Convert the mass to grams: Mass of ice = 9.17 kg × 1000 g/kg \[Mass\:of\:ice = 9,170\:g\] The energy input in 12 hours is the same as before: Energy input = 7,257,600 J Now, we can use the specific heat capacity formula to find the final temperature (T_final) of the ice. Energy input = Mass x c x (T_final - T_initial) 7,257,600 J = 9,170 g x 2.032 J/g°C x (T_final - (-5°C)) \[7,257,600 = 18,635(T_{final}+5)\] Divide both sides by 18,635: \(T_{final}+5 = \frac{7,257,600}{18,635}\) \(T_{final}+5 \approx 389.4\) Now, subtract 5 from both sides to find the final temperature: \[T_{final} \approx 384.4^{\circ}C\] The final temperature of the ice after being in sunlight for 12 hours, assuming no phase changes and uniform penetration to a depth of 1.00 cm, is approximately 384.4°C.