# Question c1015

Sep 20, 2015

${50}^{\circ} \text{C}$

#### Explanation:

In order to solve this problem, you need to know water's specific heat, which expresses the amount of heat needed to raise the temperature of $\text{1 g}$ of water by ${1}^{\circ} \text{C}$.

c_"water" = 4.18"J"/("g" ^@"C")

Now, the amount of heat is given to you in calories, which means that you're going to have to convert it to Joules

1000color(red)(cancel(color(black)("cal"))) * "4.18400 J"/(1color(red)(cancel(color(black)("cal")))) = "4184.0 J"

The equation that establishes a relationship between heat added/removed and increase/decrease in temperature looks like this

$q = m \cdot c \cdot \Delta T \text{ }$, where

$q$ - heat absorbed/removed;
$m$ - the mass of the substance;
$c$ - its specific heat;
$\Delta T$ - the change in temperature, defined as the difference between the final temperature and the initial temperature.

Rearrange the equation to solve for $\Delta T$

$\Delta T = \frac{q}{m \cdot c}$

DeltaT = (4184.0color(red)(cancel(color(black)("J"))))/(100color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ^@"C")) = 10.0 ""^@"C"

This means that the final temperature of the water will be

$\Delta T = {T}_{\text{final" - T_"initial}}$

T_"final" = "DeltaT" + T_"initial" = 10""^@"C" + 40""^@"C" = color(green)(50""^@"C")#