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

Answer:

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.