Question

if 25 grams of liquid water releases 75 J of heat when water cools from 88 degrees Celsius what is the final temperature?

Answer 1

Given :

Mass of liquid water,

`\begin{gathered} m=25\text{ g} \\ =0.025\text{ kg} \end{gathered}`

Initial temperature,

`T_i=88^(\circ)C`

Heat released,

`Q=-75\text{ J}`

The heat released is given by the formula,

`Q=mC(T_f-T_i)`

Here, C is the specific heat of water.

Rearranging the equation in order to get the final temperature;

`T_F=(Q)/(mC)+T_i`

Substituting all known values,

`\begin{gathered} T_F=(-75)/(0.025*4200)+88^(\circ)C \\ =-0.714^(\circ)C+88^(\circ)C \\ =87.286^(\circ)C \end{gathered}`

Therefore, the final temperature of the liquid water is 87.286°C.