A 251 g strip of glass wool is used to insulate a reaction flask. During the reaction the temperature of the glass wool increases
from room temperature, 22.8 °C, to 75.9 °C. Calculate the heat absorbed by the granite given that the specific heat capacity of
granite is 0.67 J/(g*°C).
3.8 kJ
8.9 kJ
13 kJ
17 kJ

8.9 KJ

Explanation:

Given data:

Mass of strip = 251 g

Initial temperature = 22.8 °C

Final temperature = 75.9 °C

Specific heat  capacity of granite = 0.67 j/g.°C

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

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 = 75.9 °C - 22.8 °C

ΔT = 53.1 °C

Q = 251 g × 0.67 j/g.°C × 53.1 °C

Q = 8929.8 J

Jolue to KJ.

8929.8J ×1 KJ / 1000 J

8.9 KJ