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 dа ta:

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