# What is the final temperature of water if 380 grams of water at 36°C absorb 47880 joules of energy?

Nov 17, 2015

${66}^{\circ} \text{C}$

#### Explanation:

In order to be able to solve this problem, you need to know the value of water's specific heat, which is usually given to be 4.18"J"/("g" ""^@"C").

A substance's specific heat tells you how much heat is required to increase the temperature of $\text{1.0 g}$ of that substance by ${1}^{\circ} \text{C}$.

Now, the equation that establishes a relationship between heat lost or gained and change in temperature looks like this

$\textcolor{b l u e}{q = m \cdot c \cdot \Delta T} \text{ }$, where

$q$ - the amount of heat
$m$ - the mass of the sample
$c$ - its specific heat
$\Delta T$ - the change in temperature, defined s final temperature minus initial temperature

So, in order to find the final temperature of the water sample, you need to first find the change in temperature, $\Delta T$. Since heat was absorbed by the water, you must have a positive value for $\Delta T$.

Plug in your values into the above equation to get

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

DeltaT = (47880color(red)(cancel(color(black)("J"))))/(380color(red)(cancel(color(black)("g"))) * 4.18color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("g"))) ""^@"C")) = 30.14^@"C"

Since the change in temperature is defined as the difference between the final temperature and the initial temperature, you can say that

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

T_"final" = 36^@"C" + 30.14^@"C" = color(green)(66^@"C")

The answer is rounded to two sig figs, the number of sig figs you have for the initial temperature and the mass of the sample.