Physics
PearlSheffler
20

Please Show all your work and the answer.

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(1) Answers
Llama11

Room temperature, θ₁ = 25°C Melting point of Lead, θ₂ = 327.5°C Specific heat capacity of Lead, c = 130 J/kg Specific latent heat of fusion of Lead, L = 2.04 * 10⁴  J/kg Mass of lead, m = 2 kg. 1) Heat transferred from room temperature to melting point Formula for specific heat Q = mc(θ₂ - θ₁)                                            Q = 2*130* (327.5 - 25) = 78650 J 2) Heat transferred to melt all of the solid at melting point Formula is for Latent heat, H = mL                                              H = 2 * 2.04*10⁴ =  4.08 * 10⁴ J = 40800J Total heat transferred =  stage 1 + stage 2 = 78650 + 40800 = 119450 Total heat transferred = 119450 J.

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