Answer:
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
