Find the initial temperature of Aluminum if 35 g of it produces 781.2 Joules at the final temperature of 44.3 ºC.

T initial =  19.55°C

Explanation:

Given data:

Mass of Al = 35 g

Energy produces = 781.2 J

Final temperature = 44.3°C

Initial temperature = ?

Solution:

Specific heat capacity:

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

Specific heat capacity of Al is 0.902 J/g.°C

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 = Final temperature - initial temperature

Q = m.c. ΔT

781.2 J = 35 g×0.902 J/g.°C× (T final - T initial)

781.2 J = 35 g×0.902 J/g.°C× (44.3°C -T initial )

781.2 J = 31.57 j/°C× (44.3°C -T initial )

781.2 J/ 31.57 j/°C = (44.3°C -T initial )

24.75°C = (44.3°C -T initial )

24.75° - 44.3°C = -T initial

T initial =  19.55°C

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3