Question

HMN Item 1921

Part A

A 2.52 g lead weight, initially at 11.0 ∘C, is submerged in 8.15 g of water at 52.3 ∘C in an insulated container.

What is the final temperature of both the weight and the water at thermal equilibrium?

Part B

A silver block, initially at 60.0 ∘C, is submerged into 100.0 g of water at 25.1 ∘C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 27.3 ∘C.

What is the mass of the silver block?

Answer 1

Part A :-

Heat gained by lead = Heat lost by water

                     mcdt = mc'dt'

Where

m = mass of lead = 2.52 g

c = specific heat capacity of lead = 0.160 J/g^oC

dt = change in temperature of lead = (t-11.0) oC             (t is the common temperature attained )

m' = mass of water = 8.15 g

c' = specific heat capacity of water = 4.184 J/g^oC

dt' = change in temperature of lead = (52.3 - t)1 ^oC     

Plug the values we get

2.52x0.160x(t-11.0) = 8.15x4.184x(52.3-t)

   t = 51.8 ^oC

Therefore the final temperature is 51.8 ^oC

Part B:-

Heat lost by silver = Heat gained by water

                     mcdt = mc'dt'

Where

m = mass of silver = ?

c = specific heat capacity of silver = 0.240 J/g^oC

dt = change in temperature of lead = 60.0-27.3 = 32.7 ^oC          

m' = mass of water = 100.0 g

c' = specific heat capacity of water = 4.184 J/g^oC

dt' = change in temperature of lead = 27.3-25.1 = 2.2 ^oC     

Plug the values we get

m = (m'c'dt') / (cdt)

   = 117.3 g

Therfore the mass of silver is 117.3 g