If 10.5 g of iron, at 25°C, absorbs 128 J of heat, what will be the final temperature of metal? (The specific heat of iron is 0.449 J g-1 °C-1)

27.5°C = Final Temperature of the metal

Explanation:

The change in temperature using the specific heat of a material could be obtained using the formula:

Q = m*S*ΔT

Where Q is heat absorbed in Joules = 128J in the problem

m is the mass of the substance = 10.5g

S is Specific Heat of the substance = 0.449J/g°C for Iron

ΔT is change in temperature = Final T - Initial T

Replacing:

128J = 10.5g*0.449J/g°C*ΔT

2.5°C = ΔT

2.5°C = Final T - 25°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