A 2.77 g lead weight, initially at 11.1 oC, is submerged in 7.94 g of water at 52.8 oC in an insulated container. clead = 0.128 J/goC; cwater = 4.18 J/goC. What is the final temperature of both the weight and the water at thermal equilibrium?

52.36 ºC

Explanation:

By the law of the conservation of energy, the heat lost by the water must equal the heat gained by lead.

q H₂O = q Pb

Now q is given by m x c x ΔT

where m is the mass, c the specific heat, and ΔT the change in temperature.

So plugging our values in the eaquation above:

q H₂O = q Pb

7.94 g x 4.18 J/gºC x (  52.8 ºC- T₂) = 2.77 g x 0.128 J/gºC x ( T₂ - 11.1 ºC )

1752.390 - 33.189 T₂  = 0.355 T₂ - 3.936

33.544 T₂ = 1756.326

T₂ = 52.36 ºC

Notice that the change in temperature for water is very small, a reflection of the very large specific heat  of water.

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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