A 1424 gram sample of a liquid at an initial temperature of 30.0 degrees C absorbs 1560 J of heat. Given the specific heat of 2.44 J/g degree C, what is the final temperature of the liquid? degree C

THE FINAL TEMPERATURE OF THE LIQUID SAMPLE  IS 30.45 DEGREE CELSIUS

Explanation:

Mass of the liquid sample = 1424 g

Initail temperature = 30 degree C

Heat evolved = 1560 J

Specific heat of the liquid = 2.44 J/g degree C

Final temperature = unknown

Since the heat evolved by a substance is the product of the mass, specific heat capacity and the change in temperature of the sample

Heat = Mass * Specific heat * change in temp.

                                       H = m c (T2-T1)

Re-arranging the formula by making T2 (final temperature) the subject of the equation, we have:

         T2= H/ m c + T1

So therefore, introducing the value of the variables and solving for T2, we have:

T2 = 1560 / 1424 * 2.44 + 30

T2 = 1560 / 3474.56 + 30

T2 = 0.4487 + 30

T2 = 30.4487 degree C

The final temperature of the liquid sample is approximately 30.45 degree C