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.

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.

The energy required to heat 500 g of  lead from 25 °C to 75 °C is 3325 J

From the question given above, the following data were obtained:

Mass (M) = 500 g

Initial temperature (T₁) = 25 °C

Final temperature (T₂) = 75 °C

Specific heat of lead (C) = 0.129 J/g°C

Initial temperature (T₁) = 25 °C

Final temperature (T₂) = 75 °C

Change in temperature (ΔT) =?

ΔT = T₂ - T₁

ΔT=  75 - 25

Mass (M) = 500 g

Change in temperature (ΔT) = 50 °C

Specific heat of lead (C) = 0.129 J/g°C

Q = MCΔT

Q = 500 × 0.129 × 50

Therefore, the energy required to heat the lead is 3325 J

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