What would be the final temperature of the system if 27.4 g of lead at 120 ◦C is dropped into 22.4 g of water at 12.4 ◦C in an insulated container? The specific heat of lead is 0.128 J/g ◦C.

Answer 1

q = m c ΔT

Before we start the calculations, we need to understand the heat transfer that is taking place between lead and water. Lead was at an initial temperature of 120 °C and then placed into the water at 12.4 °C. Heat was lost by lead and absorbed by water. Therefore, the temperature of lead should decrease whereas the temperature of water should increase. According to the Law of Conservation of energy, the above equation translates to:

- q_lead = + q_water

q_lead is negative because heat was lost, whereas q_water is positive because heat was absorbed. Substitute mcΔT on both sides of the equation, you get

- (m c ΔT)_lead = (m c ΔT)_water

Now let's plug in the numbers and use algebra to solve for the final temperature (T_f).

-27.4 g x 0.128 J g^-1 °C^-1 x (T_f - T_i)_lead = 22.4 g x 4.184 J g^-1 °C^-1 x (T_f - T_i)_water

-3.51 (T_f - 120) = 93.7 (T_f - 12.4)

-3.51 T_f + 421 = 93.7 T_f - 1162

93.7 T_f + 3.51 T_f = 1162 + 421

97.2 T_f = 1583

T_f = 1583 / 97.2 = 16.3 °C