Answer
$49.6g$
Work Step by Step
To find the mass we are going to use the equation $q= m\times C_{s}\times\Delta T$ where $q$ is the heat in joules, $m$ is the mass in grams, $C_{s}$ is the specific heat (found on table on pg. 257), and $\Delta T$ is the change in temperature in $^{\circ}C$.
To find $\Delta T$ we must subtract the final temperature from the initial temperature:
$32.4-23.0=9.4^{\circ}C$
Now we plug every thing into the equation. The specific heat of water is $4.18$.
$1.95\times10^{3}J= m\times 4.18\frac{J}{g \times^{\circ}C}\times\ 9.4^{\circ}C$
Rearranging we get:
$\frac{1.95\times10^{3}}{9.4\times 4.18}g=m$ so the mass of the substance is $49.6g$
