What is the total amount of heat absorbed by 100.0 grams of water when the temperature of the water is increased from 30.0°C to 45.0°C?

(1) 418 J (3) 12 500 J

(2) 6270 J (4) 18 800 J

The total amount of heat is the same as the energy. Because no phase change is occurring, use the formula Q = mc (delta) T. Q is the energy, m is the mass, c is the specific heat capacity of water as a liquid, and (delta) T is the change in temperature. So: Q = (100.0 g) (4.186 J/g°C) (15°C); Q = 6279 Joules.