Question

A 2.28 g lead weight, initially at 11.1 ∘C, is submerged in 8.15 gof water at 52.8 ∘C in an insulated container. What is the final temperature (degrees Celsius) of both the weight and the water at thermal equilibrium?

Answer 1

m(lead) = 2.28 g

T(lead) = 11.1 oC

C(lead) = 0.128 J/goC

m(water) = 8.15 g

T(water) = 52.8 oC

C(water) = 4.184 J/goC

T = 24.3 oC

We will be using heat conservation equation

Let the final temperature be T oC

use:

heat lost by water = heat gained by lead

m(water)*C(water)*(T(water)-T) = m(lead)*C(lead)*(T-T(lead))

8.15*4.184*(52.8-T) = 2.28*0.128*(T-11.1)

34.0996*(52.8-T) = 0.2918*(T-11.1)

1800.4589 - 34.0996*T = 0.2918*T - 3.2394

T= 52.4461 oC

Answer: 52.4 oC