What is the final temperature of both the weight and the water at thermal equilibrium? Express the temperature in Celsius to three significant figures.
A 2.80 g lead weight, initially at 10.3 ∘C, is submerged in 8.23 g of water at 52.2 ∘C in an insulated container.
A 2.80 g lead weight, initially at 10.3 ∘C, is submerged in 8.23 g of water at 52.2 ∘C in an insulated container.
1 Answer
Well, it ought to be closer to water. Why would we be incorrect if we had gotten something higher than
Looking these up... the specific heat capacities are needed.
#C_(Pb) = "0.128 J/g"^@ "C"#
#C_(w) = "4.184 J/g"^@ "C"#
By conservation of energy,
#q_(Pb) + q_w = 0#
Thus,
#q_(Pb) = -q_w#
or
#m_(Pb)C_(Pb)DeltaT_(Pb) = -m_w C_w DeltaT_w# where
#m# is the mass of something in#"g"# and#DeltaT# is its change in temperature.
We expect the cold metal to cool down the solution by absorbing heat from the water. Since we want the final temperature, and everything in this system will reach the same final temperature
#m_(Pb)C_(Pb)(T_f - T_(Pb)) = -m_w C_w (T_f - T_w)#
Most people would prefer to plug numbers in at this point.
#"2.80 g" cdot "0.128 J/g"^@ "C" cdot (T_f - 10.3^@ "C") = -"8.23 g" cdot "4.184 J/g"^@ "C" cdot (T_f - 52.2^@ "C")#
Omitting the units for simplicity, and we can put them back later...
#0.3584 (T_f - 10.3) = -34.43 (T_f - 52.2)#
#0.3584T_f - 3.692 = -34.43T_f + 1797.25#
And thus, the final temperature is...
#color(blue)(T_f) = ((1797.25 + 3.692)/(0.3584 + 34.43))""^@ "C"#
#= color(blue)(51.8^@ "C")#
And this is reasonable. It must be close to water, as it has the higher specific heat capacity... lead has such a low specific heat capacity that it cannot store as much thermal energy as water can.
