Question

a 225g sample of aluminum was heated from 22.5C to 125.5C. the specific heat of aluminum is 0.900 J/G C. how much heat was absorbed by the metal

Answer 1

Q = 20857.5 J

Step-by-step 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