Answer:
Specific heat of aluminum = 0.27 j/g.°C
Explanation:
Given data:
Mass of Al = 100 g
Initial temperature = 108.4°C
Final temperature = 68.2°C
Heat released = -1080 j
Specific heat of aluminum = ?
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 = 68.2°C - 108.4°C
ΔT = - 40.2°C
Q = m.c. ΔT
-1080 j = 100 g ×c ×- 40.2°C
-1080 j = -4020 g.°C ×c
c = -1080 j/-4020 g.°C
c = 0.27 j/g.°C
