An object with a mass of #100 g# is dropped into #500 mL# of water at #0^@C#. If the object cools by #20 ^@C# and the water warms by #4 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Jul 28, 2017

Specific heat of the object is #1#

Explanation:

In such cases, heat lost is equal to heat gained. Further heat lost / gained is given by #mst#, where #m# is the mass of the object, #s# is its specific heat and #t# is rise / fall in temperature.

Hence let #s# be the specific heat of the object, then heat lost by the object is #100xxsxx20=2000s# calories

annd heat gained by water is #500xx1xx4=2000# calories.

Hence #2000s=2000# or #s=1#