Question

A block of aluminum weighing 100g is cooled from 108.4°C to 68.2°C

with the release of 1080 joules of heat. From this data, calculate the
specific heat of aluminum

Answer 1

Specific heat of aluminum = 0.27 j/g.°C

Step-by-step 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