Answer
The mass of the boiling water that must be added to attain a final temperature of $75^{^{\circ}}C$ is $1.95$ $ kilograms$
Work Step by Step
$Q_{Before}$ = $Q_{After}$
$mC\Delta T_{1}$ $ = $ $ mC\Delta T_{2}$
$=$$(0.750 kg)C(75-10^{^{\circ}}C$) = $m_{After}C(75-100^{^{\circ}}C$)
*The specific heat capacity is the same on both sides and thus cancels out of the equation. Therefore by rearranging the equation and solving for the mass after, it is evident that $1.95 kg $ of boiling water is required.
