Answer:
The energy required is 3225 Joules.
Explanation:
Given,
mass of lead cube = 500 grams
T₁ = 25°C
T₂ = 75°C
specific heat of lead = 0.129 J/g°C
Energy required to heat the lead can be found by using the formula,
Q = (mass) (ΔT) (Cp)
Here, ΔT = T₂ - T₁ = 75 - 25 = 50
Substituting the values,
Q = (500)(50)(0.129)
Q = 3225 Joules.
Therefore, energy required is 3225 J.
