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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

User Shiramy
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1 Answer

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Answer:

THE FINAL TEMPERATURE OF THE LIQUID SAMPLE IS 30.45 DEGREE CELSIUS

Step-by-step 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

User Mrunal Pagnis
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