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Answer 1 · 2014-06-07T22:16:01

The final temperature of the lead is 46.1 °C.

The formula to use is

#q = mcΔT#,

where #q# is the heat involved, #m# is the mass, #c# is the specific heat capacity, and #ΔT# is the temperature change.

#q# = 40.5 J
#m# = 15.0 g
#c# = 0.128 J·g⁻¹°C⁻¹
#ΔT# = ?

Solve the equation for #ΔT#.

#ΔT = q/(mc) = (40.5"J")/((15.0"g")(0.128"J·g⁻¹°C⁻¹"))# = 21.1 °C

#Δ T = T_2 – T_1#

#T_2 = T_1 + ΔT# = 25.0 °C + 21.1 °C = 46.1 °C