Answer:
Q = 270 Joules (2 sig. figs. as based on temperature change.)
Explanation:
Heat Transfer Equation of pure condensed phase substance => Q = mcΔT
Mixed phase (s ⇄ l melting/freezing, or l ⇄ g boiling/condensation) heat transfer equation => Q = m∙ΔHₓ; ΔHₓ = phase transition constant
Since this is a pure condensed phase (or, single phase) form of lead (Pb°(s)) and not melting/freezing or boiling/condensation, one should use
Q = m·c·ΔT
m = mass of lead = 35.0g
c = specific heat of lead = 0.16J/g°C
ΔT = Temp change = 74°C - 25°C = 49°C
Q = (35.0g)(0.16J/g·°C )(49°C) = 274.4 Joules ≅ 270 Joules (2 sig. figs. as based on temperature change.)
