# Suppose 3260 joules are absorbed by 135 grams of water. If the initial temperature of the water is 21.4°C, what is its final temperature?

Nov 25, 2017

According to the given data, the water's final temperature is approximately 27.2°C.

#### Explanation:

The specific heat of water is: (4.18J)/(g*°C)

And the equation we want to relate these data with is:

$q = m {C}_{s} \Delta T$

Hence,

3260J = 135g * (4.18J)/(g*°C) * DeltaT
$\therefore \Delta T \approx 5.78$

Our final temperature is T_f = 21.4 + 5.78 = 27.2°C