How many joules are absorbed when 83.2 grams of water is heated from 30.0 °C to 74.0 °C? 10,500 joules 12,200 joules 15,300 joules 18,200 joules

15,300 joules.

Explanation:

To solve this problem, we can use the relation:

Q = mcΔT,

Where, Q is the amount of heat needed to raise the temperature of the water (J).

m is the mass of the water (m = 83.2 g).

c is the specific heat capacity of the metal (c = 4.186 J/g.°C).

ΔT is the temperature difference (ΔT = final T - initial T = 74.0 °C - 30.0 °C = 44.0 °C).

∴ Q = mcΔT = (83.2 g) (4.186 J/g.°C) (44.0 °C) = 15324.1 J ≅ 15,300 joules.