Question

A 500 gram cube of lead is heated from 25 °C to 75 °C. How much energy was required to heat the lead? The specific heat of lead is 0.129 J/g°C.

Answer 1

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.