A sample of an unkown metal has a mass of 120.7g. As the sample cools from 90.5C to 25.7C, it releases 7020J of energy. What is the specific heat of the sample

Specific heat of metal = 0.9 j/g.°C

Explanation:

Given data:

Mass of metal = 120.7 g

Initial temperature of sample = 90.5°C

Final temperature of sample = 25.7 °C

Heat released = 7030 J

Specific heat of metal = ?

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 = final temperature - initial temperature

ΔT = 25.7 °C - 90.5 °C

ΔT =  - 64.8°C

Now we will put the values in formula.

Q = m.c. ΔT

7020 J = 120.7 g × c  ×  - 64.8°C

7020 J = 7821.36 g.°C  × c

c = 7020 J / 7821.36 g.°C

c = 0.9 j/g.°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