If 980 kJ of energy are added to 6.2 L of water at 291 K, what wil the final temperature of the water be?

1 Answer
Jun 6, 2017

328.81k

Explanation:

Use the equation q=mcDeltaT, where q is energy, m is mass, c is the specific heat capacity, and DeltaT is the change in temperature. The mass of water is 1000g per litre, so 6.2L = 6200g. The specific heat capacity of water is ~~4.18 J*g^-1*K^-1, so:
980,000 = 6,200 * 4.18 * DeltaT
DeltaT = \frac{980,000}{6,200 * 4.18} = 37.81K
:. temperature = 291 + 37.81 = 328.81K .
But realistically it would never be this high as energy will escape to the surroundings. This is only a theoretical value.