Final answer:
To warm 1.60 kg of sand from 30.0 ℃ to 100.0 ℃, 94,080 joules of heat is required, using the specific heat capacity of sand and the formula for heat transfer.
Step-by-step explanation:
To calculate how much heat is required to warm 1.60 kg of sand from 30.0 ℃ to 100.0 ℃, we use the formula for heat transfer, Q = mcΔT, where 'Q' is the heat energy, 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature. Using the given specific heat capacity of sand, which is 0.840 J/(g·℃), and converting the mass of sand to grams (1.60 kg = 1600 g), we can find the required heat:
Q = (1600 g) × (0.840 J/g·℃) × (100.0·℃ - 30.0·℃) = (1600) × (0.840) × (70) = 94,080 J
Therefore, 94,080 J of heat energy is needed to warm the sand from 30.0 ℃ to 100.0 ℃.