Question

How much energy is required to change the temperature of 35.0 g of Lead

Answer 1

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.)