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

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.