Question

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?

Answer 1

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 = mC_sDeltaT`

Hence,

`3260J = 135g * (4.18J)/(g*°C) * DeltaT`
`thereforeDeltaT approx 5.78`

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