How much energy is required to raise the temperature of a 300.0 gram block of lead from 22.3°C to 59.9°C? The specific
heat of lead is 0.129 J/gºC.
Round your answer to the nearest whole number.

Q = 1455 J

Explanation:

Given data:

Mass of block = 300 g

Initial temperature = 22.3 °C

Final temperature = 59.9°C

Specific heat of lead = 0.129 j/g.°C

Energy required = ?

Solution:

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 = T2 - T1 =  59.9°C - 22.3 °C

ΔT = 37.6  °C

Q = 300 g × 0.129 j/g.°C ×  37.6  °C

Q = 1455 J

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