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 dа ta:

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