An object with a mass of #160 g# is dropped into #880 mL# of water at #0^@C#. If the object cools by #18 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
A08
Feb 8, 2016

#s_(object)=2.dot 4# #cal//gm#

Explanation:

Assumption : Specific gravity of water at #0^@ C# is 1. (actually 0.9998)
#=># weight of #880 ml# of water is #880 gm#.

Now by Law of conservation of energy
Heat lost by object = Heat gained by water.

Also Heat gained or lost #=m*s*Delta t#
where #m# is the mass of object, #s# its specific heat and#Delta t# Change in the temperature of the object.

Heat lost by object#=160*s_(object)*18#
Heat gained by water #=880*1*8#
(Specific heat of water in cgs units is 1 calorie/gm)
Equating two we obtain
#160*s_(object)*18=880*1*8#
#s_(object)=(880*8)/(160*18)#
#s_(object)=2.dot 4# #cal//gm#

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