Q = smT

S water = 4.184 J/g℃

1 calorie = 4.184 J

How much heat, in joules and in kilojoules, is required to raise the temperature of a 33.0 g sample of Al from 25 to 68 ℃? Specific heat capacity of Al is 0.89 J/g℃.

The heat required is 1262.91 joules or 1.26 kilojoules.

Explanation:

The quantity of heat (Q) required to raise the temperature of a substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Since,

Q = ?

Mass of aluminum sample = 33.0g

C = 0.89 J/g℃

Φ = (Final temperature - Initial temperature)

= 68°C - 25°C = 43°C

Then, Q = MCΦ

Q = 33.0g x 0.89 J/g℃ x 43°C

Q = 1262.91 J

Since the heat in joules is 1262.91, obtain heat in kilojoules.

If 1000 joules = 1 kilojoules

1262.91 joules = Z

To get Z, cross multiply

Z x 1000 = 1262.91 x 1

1000Z = 1262.91

Z = 1262.91 / 1000

Z = 1.26291 kilojoules (when placed as 3 significant figures, Z is 1.26 kilojoules)

Thus, the heat required is 1262.91 joules or 1.26 kilojoules.