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?

T(final) = 328.78 K

Explanation:

Given data:

Energy added = 980 kj

Mass of water  = 6.2 L

Initial temperature = 291 K

Final temperature = ?

Solution:

Mass of water  = 6.2 L = 6.2 Kg = 6200 g

Specific heat of water = 4.184 j/g.K

Energy added = 980 kj = 980 × 1000 = 980,000 J

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

980,000 J = 6200 g × 4.184 j/g.K × T(final) - 291 K

980,000 J = 25940.8 j/K  × T(final)  - 291 K

980,000 J / 25940.8 j/K = T(final) - 291 K

T(final)  - 291 K = 37.78 K

T(final)  = 37.78 K  + 291 K

T(final)  = 328.78 K

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2