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?

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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 $1 g/(cm^3))$

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

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

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

An object with a mass of 8 g8 g is dropped into 500 mL500 mL of water at 0^@C0^@C. If the object cools by 50 ^@C50 ^@C and the water warms by 6 ^@C6 ^@C, what is the specific heat of the material that the object is made of?