Question

In: Chemistry

b) A 50.53 g sample of a metal at 290.00 C is placed in a calorimeter...

b) A 50.53 g sample of a metal at 290.00 C is placed in a calorimeter containing 150.00 g of water at 20.00 C. The temperature stopped changing at 29.42 C. What is the specific heat of the metal? What is the identity of the metal?

c) A 150.0 g sample of copper at 75.00 C is added to 150.00 g of water initially at 23.00 C Assuming that all of the heat lost by the copper is absorbed by the water, calculate the final temperature of the copper and the final temperature of the water.

Solutions

Expert Solution

b.

     Heat lose of metal                            =                             Heat gain of water

    mcT                                               =                             mcT

50.53*c*(290-29.42)                            =                            150*4.184*(29.42-20)

                                    c                        = 0.44J/g-0C

The metal is iron

c.

Heat lose of copper                              =                 Heat gain of water

mcT                                                   =                    mcT

150*0.386*(75-t)                                     =                  150*4.184*(t-23)

                                      t                             =   27.40C

The final temperature of copper and water                  = 27.40C >>>>answer

queen_honey_blossom answered 1 year ago
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