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In: Chemistry

A sample of metal with a mass of 350 g is heated to 100 degrees celcius...

A sample of metal with a mass of 350 g is heated to 100 degrees celcius and dropped into 400 g of water at 21.4 degrees celcius. The temp of the water rises to 28.0 celcius. Assume no heat is lost to the environment.

What is the amount of heat gained?

What is the specific heat of the metal?

Solutions

Expert Solution

Heat gain of water q = mCT

                                   = 400*4.184*(28-21.4)

                                   = 11045.76J

Heat lose of metal             =       Heat gain of water

mCT                             =        mCT

350*C*(100-28)              =          400*4.184*(28-21.4)

          C                          = 0.438j/g-c

    specific heat of metal = 0.438J/g-c

queen_honey_blossom answered 1 year ago
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