Final answer:
To find out how much heat is absorbed by 2.50 kg of water to increase its temperature from 10.0°C to 60.0°C, the formula Q = mcΔT is used with the given specific heat of water. After calculation, the heat absorbed is found to be approximately 523,000 J.
Step-by-step explanation:
To calculate the amount of heat absorbed by the water, we use the equation Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. The given values are: m = 2.50 kg of water, c = 4,184 J/kg°C, and ΔT = 60.0°C - 10.0°C = 50.0°C. Plugging these values into the equation, we get:
Q = (2.50 kg) × (4,184 J/kg°C) × (50.0°C) = 524,000 J.
Therefore, the correct answer is C. 523,000 J, after rounding to the nearest hundred as usually done in physics calculations.