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;

$T_F=\frac{Q}{mC}+T_i$

Substituting all known values,

$\begin{gathered} T_F=\frac{-75}{0.025\times4200}+88^{\circ}C \\ =-0.714^{\circ}C+88^{\circ}C \\ =87.286^{\circ}C \end{gathered}$

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