Question

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?

Answer 1

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.