# 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?

Jun 6, 2017

328.81k

#### Explanation:

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