# 4790 joules of heat are added to 222 grams of water originally at 27.7 C. What is the final temperature of the water?

Nov 3, 2016

If I am correct, you will need to know the equation Q=mc ΔT to solve this problem.

#### Explanation:

Here, you are given the values for Q, m and c can be obtained from your textbook. According to mine, the value should be 4.19 J/C.

4790 J= 222g * 4.19 * (x°C-27.7°C)

Note that I let x represent the final temperature. And now, all you have to do is some simple arithmetic and solve for x!

Nov 3, 2016

The final temperature is $\text{32.9"^@"C}$.

#### Explanation:

Use the specific heat equation:

Known

$Q = \text{4790 J}$
c_"water"=4.184 "J"/("g·"^@"C")
$m = \text{222 g}$

${T}_{\text{initial"="27.7"^@"C}}$

Unknown

${T}_{\text{final}}$

Solution

Rearrange the equation to isolate $\Delta T$.

$Q = c m \Delta T$

Divide $Q$ by $c m$.

$\Delta T = \frac{Q}{c m}$

Delta T=(4790cancel"J")/((4.184cancel"J")/(cancel"g·"^@"C")xx222cancel"g"$=$$\text{5.16"^@"C}$

Now that $\Delta T$ is known, we'll write it as equal to the final temperature minus the initial temperature.

$\Delta T = {T}_{f} - {T}_{i}$

Add ${T}_{1}$ to both sides.

$\Delta T + {T}_{i} = {T}_{f}$

${5.16}^{\circ} \text{C"+"27.7"^@"C"="32.9"^@"C}$