If 18.0 kJ of energy are applied to 250. grams of water at 37°C, what will be the final temperature

54.1°C

Explanation:

Now we have the following parameters from the question;

Heat applied (H) = 18.0*10^3 Joules

Mass of the water (m) = 250g = 0.25Kg

Initial temperature of the water (θ1) = 37°C

Final temperature of the water (θ2) = the unknown

Heat capacity of water (c) = 4200JKg-1

From;

H = mc (θ2-θ1)

Substituting values appropriately

18*10^3 = 0.25 * 4200 (θ2-37)

18*10^3 = 0.25 * (4200θ2 - 155400)

18*10^3 = 1050θ2 - 38850

18*10^3 + 38850 = 1050θ2

56850 = 1050θ2

θ2 = 56850/1050

θ2 = 54.1°C