An insulated container is used to hold 44.60 g of water at 26.20 degree C. A sam

Question

An insulated container is used to hold 44.60 g of water at 26.20 degree C. A sample of copper weighing 16.90 g is placed in a dry test tube and heated for 30 minutes in a boiling water bath at 100.10 degree C. The heated test tube is carefully removed from the water bath with laboratory tongs and inclined so that the copper slides into the water in the insulated container. Given that the specific heat of solid copper is 0.385 J/(g middot degree C), calculate the maximum temperature of the water in the insulated container after the copper metal is added.

Explanation / Answer

Heat lost by copper = heat gained by water

                    mcdt = m'c'dt'

Where

m = mass of copper = 16.90 g

c = specific heat capacity of copper = 0.385 J/goC

dt = change in temperature = final - initial = (100.10 - t ) oC     t = common temperature attained by the system

m' = mass of water = 44.60 g

c' = specific heat capacity of water = 4.18 J/goC

dt' = change in temperature of water = final - initial = (t - 26.20) oC

Plug the values we get

16.90 x 0.385 x (100.10-t) = 44.60 x 4.18x (t-26.20)

(100.10-t) = 28.65 x(t-26.20)

100.10 - t = 28.65 t - 750.7

29.65 t = 850.8

         t = 28.69 oC

Therefore the maximum temperature of water is 28.69 oC

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