Final answer:
By converting 3500 calories to joules and using the heat equation (q = m × c × ΔT) with the specific heat capacity of water, it's determined that approximately 63.5 grams of water can be heated from 20.0°C to 75.0°C. None of the options provided in the question is correct.
Step-by-step explanation:
To determine how many grams of water can be heated from 20.0°C to 75.0°C using 3500 calories, we should first convert calories to joules since the specific heat capacity of water is usually given in joules per gram degree Celsius (J/g°C). Knowing that 1 calorie equals 4.184 joules, we can calculate the total energy in joules:
3500 cal × 4.184 J/cal = 14644 J
Next, we can use the formula q = m × c × ΔT, where q is the heat in joules, m is the mass in grams, c is the specific heat capacity (4.184 J/g°C for water), and ΔT is the temperature change in degrees Celsius.
Rearranging the formula to solve for the mass m gives us:
m = q / (c × ΔT)
Then substituting the known values:
m = 14644 J / (4.184 J/g°C × (75°C - 20°C))
m = 14644 J / (4.184 J/g°C × 55°C)
m = 14644 J / 230.62 J/g = 63.5 g (approximately)
Therefore, using 3500 calories, you can heat approximately 63.5 grams of water from 20.0°C to 75.0°C, which means none of the options (a, b, c, or d) listed in the question accurately reflect the correct answer.