Question

A lead pellet with a mass of 26.47 g and at an initial temperature of 89.98 °C was dropped into a beaker containing 100.0 g of water at an initial temperature of 22.50 °C. Given the sp heat capacity of water = 4.184 J/gK and of lead = 0.158 J/gK, calculate the final temperature of both the lead and the water?

Answer 1

Lead pellet;

mass, m1 = 26.47 g

specific heat capacity of lead, s1= 0.158 J/gK

initial temperature, t1= 89.98 °C

Let t be the temperature lost by lead pellet.

Water:

mass, m2 = 100.0 g

specific heat capacity of water, s2= 4.184 J/gK

initial temperature, t2= 22.50 °C

Let t be the temperature gained by water.

We know that q = ms dt

Heat lost by lead pellet = Heat gained by water

Then, m1 s1 (t1-t) = m2 s2 (t-t2)

substitute all the known values,

26.47 x 0.158 x (89.98 -t) = 100 x 4.184 (t-22.5)

26.47 x 0.158 x 89.98 - 26.47 x 0.158 x t = 100 x 4.184 x t - 100 x 4.184 x 22.5

100 x 4.184 x t + 26.47 x 0.158 x t = 26.47 x 0.158 x 89.98 + 100 x 4.184 x 22.5

[(100 x 4.184) +(26.47 x 0.158) ] t =  [(26.47 x 0.158 x 89.98) + (100 x 4.184 x 22.5)]

t = [(26.47 x 0.158 x 89.98) + (100 x 4.184 x 22.5)] / [(100 x 4.184) + (26.47 x 0.158) ]

= (376.32 + 9414 ) / (418.4 + 4.182)

= 9790.32 / 422.582

= 23.17 °C

t = 23.17 °C

Therefore, final temperature of lead = t1 -t = 89.98 °C - 23.17 °C = 66.81 °C

final temperature of water = t2 + t = 23.17 °C + 22.5 °C = 45.67 °C

Hence, final  temperature of lead = 66.81 °C

   final temperature of water = 45.67 °C