Final answer:
To heat water from 20.0°C to 65.0°C using
13500.0 Joules, you would need approximately 76.71 grams of water.
Step-by-step explanation:
To determine how many grams of water can be heated from 20.0°C to 65.0°C using
13500.0 Joules of energy, we can use the equation:
Energy = mass × specific heat capacity temperature change
Rearranging the equation to solve for mass: mass = Energy / (specific heat capacity × temperature change)
Inserting the given values:
mass = 13500.0 J / (4.184 J/g°C × (65.0°C - 20.0°C))
Calculating the mass:
mass = 13500.0 J / (4.184 J/g°C × 45.0°C)
mass = 76.71 grams