An insultaed container holds 40.30g of water at 23.10 celcius.Sample of copper weighing 7.50g is placed...

An insultaed container holds 40.30g of water at 23.10 celcius.Sample of copper weighing 7.50g is placed in a dry test tube and heated for 30 min in boiling water at 100.10 celsius. given specific heat of copper is .385 , calculate max temp of water in container after copper metal is added.

Expert Solution

The heat released by copper = heat gained by water

mcdt = m'c'dt'

Let t be the common temperature

Where

m = mass of copper = 7.50 g

c = specific heat of copper = 0.385J/(goC)

dt = change in temperature of copper = 100.10 - t

m' = mass of water = 40.30 g

c' = specific heat of water = 4.186 J/(goC)

dt' = change in temperature = t - 23.10

Plug the values we get

7.50x0.385 x ( 100.10-t) = 40.30x4.186x(t-23.10)

289.0 - 2.887t = 168.7t - 3896.9

t = 24.4 ^oC

Therefore the max temp of water in container after copper metal is added is 24.4 ^oC

queen_honey_blossom answered 2 years ago

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