Answer:
Q = 20857.5 J
Explanation:
Given data:
Mass of aluminum = 225 g
Initial temperature = 22.5°C
Final temperature = 125.5°C
Specific heat of Al = 0.900 j/g.°C
Heat absorbed by metal = ?
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 = 125.5°C - 22.5°C
ΔT = 103°C
Now we will put the values in formula.
Q = 225 g × 0.900 j/g.°C ×103°C
Q = 20857.5 J
