A 75 g sample of a substance is placed in boiling water until its temperature reaches 100 C. It is then immediately transferred to a calorimeter containing 100 g of water at 24.4 C. The final temperature of the sample and the water is 34.9 C. What is the specific heat of the sample?

specific heat of sample = 0.898 j/g.°C

Explanation:

Given data:

Mass of sample = 75 g

Initial temperature of water = 100°C

Mass of water = 100 g

Initial temperature of water = 24.4°C

Final temperature of sample ans water = 34.9°C

Specific heat of sample = ?

Solution:

ΔT(water) = 34.9°C - 24.4°C

ΔT(water) = 10.5°C

ΔT(sample) = 34.9°C - 100°C

ΔT(sample) = - 65.1°C

-Q(sample) = Q(water)

-m×c×ΔT   =  m×c×ΔT

-[75 g × c × - 65.1°C] = 100 g × 4.18 j/g.°C × 10.5°C

-[-4882.5 g.°C × c = 4389 j

4882.5 g.°C × c = 4389 j

c =  4389 j/ 4882.5 g.°C

c = 0.898 j/g.°C

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