What is the total calories lost calculated from the cooling of 500 grams of water?

To understand how to calculate the total calories lost from the cooling of 500 grams of water, it's important to recall that the specific heat capacity of water is approximately 1 calorie/gram °C. This means that it takes 1 calorie to raise the temperature of 1 gram of water by 1 degree Celsius.

If we know the temperature change (let's denote it as ΔT) that the water undergoes, we can calculate the total calories lost using the formula:

Total calories lost = mass (in grams) × specific heat capacity × temperature change (in °C)

In this scenario, if the calculation has been performed and the result yields 12,500 calories, it suggests that either the mass of water is significant enough in combination with the cooling temperature resulting in this amount of energy lost.

Assuming the water cools from a higher temperature down to a lower one (for instance from 100°C to 50°C, a difference of 50°C), the calculation would be:

Total calories lost = 500 grams × 1 cal/g°C × 50°C = 25,000 calories

However, if the question is set in a context where only a certain range is being measured, leading to the 12