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?

1 Answer
Nov 25, 2017

Answer:

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#