An object with a mass of #8 g# is dropped into #500 mL# of water at #0^@C#. If the object cools by #50 ^@C# and the water warms by #6 ^@C#, what is the specific heat of the material that the object is made of?

Question

1808 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-08-19T08:21:15

7.5 cal/(g*degrees C)

#mo*so*Delta(t_object) = mw*sw*Delta(tw)#

m=mass (grams)
s= specific heat
t=temperature
and for subscripts:
o stands for the object and w stands for water in the container.

You can now compute the specific heat assuming water has a density of 1 gram per cm^3 at 0 degrees Celcius.

500 mL water = 500 grams water (density is #1 g/(cm^3))#

#8*so*50=500*1*6#

#so = (500*6)/(50*8)#

#so=7.5 (cal)/(g*degrees C)#