Question

if 980 KJ of energy are added to 6.2 L of water at 291 K, what will the final temperature of the water be

Answer 1

The final temperature of the water will be 328.81 K .

Step-by-step explanation:

Using the equation, q = mcΔT

here, q = energy

m= mass

c= specific heat capacity

ΔT= change in temperature

Mass of water = 1kg (1000 g ) per liter

∴ 6.2 Liter of water = 6200 g

c of water ≈ 4.18 J /g/K

Now,

980000 = 6200*4.18*ΔT

ΔT = 37.81 K

∴ final temperature of the water = 291 + 37.81 = 328.81 K