Final answer:
To heat 19.0 g of water from 31.0°C to 55°C, it would take approximately 456 calories, with the closest given option being 570 calories.
Step-by-step explanation:
To determine how many calories of heat are needed to heat 19.0 g of water from 31.0°C to 55°C, we use the formula:
Q = mcΔT
Where Q is the heat energy, m is the mass of water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is 1 cal/g°C, which is equivalent to 4.184 J/g°C. By substituting the values into the formula, we get:
Q = 19.0 g × 1 cal/g°C × (55°C - 31°C)
Q = 19.0 g × 1 cal/g°C × 24°C
Q = 456 cal
The correct answer is not listed explicitly, so you would need to use standard rounding rules if these were the only options provided. Since 456 cal is closest to Option 2: 570 calories, one might choose that in a multiple-choice setting unless instructed otherwise on how to handle numbers that do not match the options exactly.