Question

Determine the specific heat of a material if a 35-g sample absorbed 96 J as it was heated from 293 K to 313 K.

Answer 1

The specific heat capacity of material is 0.14 j/g.K.

Explanation:

Given data:

Mass of sample = 35 g

Heat absorbed = 96 j

Initial temperature = 293 K

Final temperature = 313 K

Specific heat capacity = ?

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 =  313 K - 293 K = 20 K

96 J = 35 g × c × 20K

96 J = 700 g/K × c

c = 96 J/700 g/K

c = 0.14 j/g.K

Answer 2

The specific heat of a material is 0.137 J/g°C

Explanation:

The specific heat is given by the formula:

c = q/(m × ΔT)

Where,

q = Heat in joules = 96 J (given)

m = Mass of the material = 35 g

ΔT = Change in temperature = T₂ - T₁

Now, the formula becomes,

c = q/(m × (T₂ - T₁))

T₁ = 293 - 273.15 = 19.85°C

T₂ = 313 - 273.15 = 39.85°C

On substituting the values, we get,

c = 96/(35 × (39.85 - 19.85))

c = 96/(35 × (20))

c = 96/700

∴ c = 0.137 J/g°C