Question

Question #ac941

Answer 1

`t=7ºC`

Explanation:

The medium heat capacity of liquid water is 4.187 kJ/kgK, and of ice is 2.108 kJ/kgK, and heat of melting is 334 kJ/kg.

http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

Considering -10ºC as our zero energy., the water at 30ºC will have the total energy of heating from -10ºC to 0ºC, the melting and the heating from 0ºC to 30ºC

`20 g xx[(2.108 J/(gK))xx10K+334J/g+(4.187 J/(gK))xx30K]=9614 J`.

Now this energy, must be distributed by a total of 25 g.

First we heat from 10ºC to 0ºC:

`25 g xx(2.108 J/(gK))*10K=527J`

It remains `9614-527=9087J`

For melting the ice we have

`25 g xx334 J/g=8350 J`

It remains `9087-8350=737 J`

This remaining energy will heat the water from 0ºC:

`25 g xx(4.187 J/gK)*(t/(ºC))=737 J`

`t=7ºC`