Answer:
Specific heat of substance = 0.897 J/g.°C
Explanation:
Given data:
Specific heat of substance = ?
Mass of substance = 25.0 g
Heat absorbed = 493.4 J
Initial temperature = 12.0 °C
Final temperature = 34°C
Solution:
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 = 34°C -12.0°C
ΔT = 22°C
493.4 J = 25 g ×c× 22°C
493.4 J = 550 g.°C ×c
c = 493.4 J / 550 g.°C
c = 0.897 J/g.°C
