Final answer:
When 6000 calories are added to 10.0 grams of water, the calculation implies a temperature rise of 600°C. However, since water boils at 100°C and will not exceed that temperature in liquid form, the final temperature will be 100.0°C, represented by option A.
Step-by-step explanation:
If 6000 calories of energy are added to 10.0 grams of water starting at 50.0°C, the final temperature of the water can be determined by using the concept of specific heat capacity. The specific heat capacity of water is 1.00 cal/g°C, which means that it takes 1 calorie to raise the temperature of 1 gram of water by 1°C. Given this information, and the fact that we have 10.0 grams of water, the temperature increase can be found by dividing the energy added by the product of the specific heat capacity and the mass of water.
To calculate the temperature increase: Temperature increase (in °C) = energy added (cal) / (mass of water (g) × specific heat capacity (cal/g°C)) = 6000 cal / (10.0 g × 1.00 cal/g°C) = 600°C.
However, since water boils at 100°C and cannot become hotter than 100°C at normal atmospheric pressure while remaining in the liquid state, the final temperature of the water will not actually reach 600°C after adding 6000 calories. Instead, the water will reach its boiling point and then start to change state from liquid to gas (boil) at that temperature. The final temperature of the water, therefore, will be 100.0°C, which corresponds to option A.