Final answer:
To melt a 1.8 cm thick block of ice absorbing 100 W/m² of sunlight for 12 hours a day, takes approximately 0.0035 years, roughly 1.28 days. The calculation involves determining the energy required to melt the ice and dividing it by the energy absorbed daily from the sunlight.
Step-by-step explanation:
To determine how long it takes to melt a block of ice with sunlight, we first need to calculate the total energy required to melt the ice. The heat transfer Q required to melt a mass m of ice at 0°C can be found using the formula Q = mLf, where Lf is the latent heat of fusion of ice (approximately 334,000 J/kg). Once we have the heat transfer value, we can determine the time t required to melt the ice by dividing Q by the power-absorbed P.
Firstly, we calculate the mass m of the ice from its volume V and the density of ice (ρ ≈ 920 kg/m³). The volume V = area × thickness = 1.0 m² × 0.018 m.
Therefore, m = ρV.
Since the ice absorbs 100 W/m², and considering the area is 1.0 m², the power absorbed P is 100 Watts. The energy per day is P multiplied by the number of seconds in the sunlight hours (12 hours × 3600 seconds/hour). To find the time in years t, divide the total energy required Q by the daily energy absorption and then by the number of days in a year.
Calculating the details: The volume of the ice V = 1.0 m² × 0.018 m = 0.018 m³. The mass m = 920 kg/m³ × 0.018 m³ = 16.56 kg. The total energy Q required to melt the ice is Q = mLf = 16.56 kg × 334,000 J/kg = 5.531 x 106 J. The daily energy absorption is 100 W × 12 hours × 3600 s/hour = 4.32 x 106 J/day.
The time in days t = Q / daily energy absorption = 5.531 x 106 J / 4.32 x 106 J/day ≈ 1.28 days. Since we need the time in years, we divide by 365.25 (accounting for leap years), so t ≈ 1.28 days / 365.25 days/year ≈ 0.0035 years.