# If 24500 J is applied to 125g of water at 35 C, what will the final temperature of the water be?

May 4, 2017

81.9C

#### Explanation:

$\delta H = m \times C p \times \delta T$, where $\delta H$ is the enthalpy or change in energy, m is the mass, Cp is the specific heat, and $\delta T$ is the change in temperature.

We know $\delta H$= 24500J, the specific heat of water is 4.18J/mol*k, and m=125g

So we alter our formula a little bit and substitute in what we know:

$\delta T = \frac{\delta H}{m \times C p} = \frac{24500 J}{125 g \times 4.18 \frac{J}{m o l \times k}} = 46.9 k$

Celsius and kelvin have the same scale for units, so simply add the change to the original temperature.

We have: Tfinal = 35+46.9=81.9C

Hope that helps!