A 22.5 g sample of an unknown material is heated from 25oC to 90oC. If 431 J of heat energy is absorbed in the heating process, what is the specific heat of the material?

c = 0.29 J/g.°C

Explanation:

Given data:

Mass of sample = 22.5 g

Initial temperature = 25°C

Final temperature = 90°C

Heat absorbed = 431 J

Specific heat of material = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 90°C - 25°C

ΔT = 65°C

431 J = 22.5 g × c × 65°C

431 J = 1462.5 g.°C  × c

c = 431 J/1462.5 g.°C

c = 0.29 J/g.°C

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