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

1 Answer
May 4, 2017



#delta H = mxxCpxxdeltaT#, 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:

#deltaT=(deltaH)/(mxxCp)=(24500J)/(125gxx4.18J/(molxxk)) = 46.9k#

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!