An object with a mass of #60 g# is dropped into #900 mL# of water at #0^@C#. If the object cools by #40 ^@C# and the water warms by #16 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Mar 6, 2016

Answer:

Specific heat of the material that the object is #6#.

Explanation:

In such cases heat lost by one equals heat gained by other. Heat lost / gained is given by product of mass #m#, specific heat #s# (which is #1# for water) and change in temperature,

Let the specific heat of the mass of #60g#, which cools by #40^oC# be #s#. Hence heat lost is #60xx40xxs# calories.

900mL of water means #900g# and it warms by 16∘C. Hence heat gained by it is #900xx16=14400# calories.

Hence #60xx40xxs=14400# or #x=14400/(60xx40)=6#

Hence, specific heat of the material that the object is #6#.