9. Calculate the specific heat capacity of titanium if a 43.56 g sample absorbs 0.476 kJ as its temperature changes from 20.5 oC to 41.2 oC.

c = 0.528 J/g.°C

Explanation:

Given data:

Mass of titanium = 43.56 g

Heat absorbed = 0.476 KJ  = 476 j

Initial temperature = 20.5°C

Final temperature = 41.2°C

Specific heat capacity = ?

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 = 41.2°C - 20.5°C

ΔT = 20.7 °C

476 J = 43.56 g × c × 20.7 °C

476 J = 901.692 g.°C × c

c = 476 J / 901.692 g.°C

c = 0.528 J/g.°C  

b.

the accuracy of the data.