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  

percent abundance for cu-63 can be shown as:

63 x + 65 (1-x) = 62.9296

65 - 62.9296 = 2 x

x = 1.035

thus, percent abundance for cu-63 is 1.035

percent abundance for cu-65 can be shown as:

63 x + 65 (1-x) = 64.9278

65 - 64.9278 = 2 x

x = 0.0361

thus, percent abundance for cu-63 is 0.0361

the mass spectrum of cu-63 showing the intense peak as, this isotope is found in nature more abundantly around 92.38 % in comparison to cu-65 which is found 62.58 % naturally