Question

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

Answer 1

54.1°C

Step-by-step 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