Final answer:
Approximately 2779.776 joules of heat energy are required to raise the temperature of 18 g of liquid water from 70°C to 107°C, using the specific heat capacity of liquid water which is 4.184 J/g°C.
Step-by-step explanation:
To calculate the amount of heat energy required to raise the temperature of 18 g of liquid water from 70°C to 107°C, you can use the equation q = mcΔT. Here, q represents the heat energy in joules, m is the mass of the water in grams, c is the specific heat capacity in J/g°C, and ΔT is the change in temperature in degrees Celsius. Given that the specific heat capacity of liquid water is 4.184 J/g°C, and we are only considering the heating of liquid water without a phase change, the formula simplifies to:
q = (18 g) × (4.184 J/g°C) × (107°C - 70°C)
When you plug the values into the equation, the calculation is as follows:
q = (18 g) × (4.184 J/g°C) × (37°C)
q = 18 × 4.184 × 37
q = 2779.776 J
Therefore, it requires approximately 2779.776 joules of heat energy to increase the temperature of 18 g of liquid water from 70°C to 107°C.