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.
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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