Determine the specific heat ofmaterial if a 12g sample absorbed 48j as it was heated from 20-40

c =0.2 J/g.°C

Explanation:

Given data:

Specific heat of material = ?

Mass of sample = 12 g

Heat absorbed = 48 J

Initial temperature = 20°C

Final temperature = 40°C

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 =  40°C -20°C

ΔT =  20°C

48 J = 12 g×c×20°C

48 J =240 g.°C×c

c = 48 J/240 g.°C

c =0.2 J/g.°C