May 17, 2017

${T}_{\text{Initial"=40.3^@ "C}}$

#### Explanation:

Solving this problem requires knowing the specific heat capacity of water. Water requires $4.184$ Joules of heat for the temperature of 1 gram of water to increase ${1}^{\circ}$ Celsius. By comparison, copper is only $0.385$.

Next, we need to use the formula

$q = m c \Delta T$

Note that the $\Delta T = {T}_{\text{Final"-T_"Initial}}$. In the problem you gave, we are given the final temperature, ${T}_{\text{Final}}$, because it says "Water...just begins to boil". Water boils at ${100}^{\circ}$ Celsius, so we have:

Final temperature, ${T}_{\text{Final"=100^@ "C}}$
Mass, $m = 2150 \text{ g}$
Specific heat capacity, $c = 4.184 \text{ ""J"//"g"*^@"C}$
Heat energy absorption, $q = 5.37 \times {10}^{5} \text{ J}$

Plugging these values in, you get:
$q = m c \left({T}_{\text{Final"-T_"Initial}}\right)$
$5.37 \times {10}^{5} = \left(2150\right) \left(4.184\right) \left(100 - {T}_{\text{Initial}}\right)$
$59.7 = 100 - {T}_{\text{Initial}}$
${T}_{\text{Initial"=100-59.7=40.3^@ "C}}$

I chose to not type all the dimensional analysis for ease of giving you the answer, but your teacher will want to see it! Also, be careful to use the correct number of significant figures.