# 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?

Aug 19, 2016

7.5 cal/(g*degrees C)

#### Explanation:

$m o \cdot s o \cdot \Delta \left({t}_{o} b j e c t\right) = m w \cdot s w \cdot \Delta \left(t w\right)$

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))

$8 \cdot s o \cdot 50 = 500 \cdot 1 \cdot 6$

$s o = \frac{500 \cdot 6}{50 \cdot 8}$

$s o = 7.5 \frac{c a l}{g \cdot \mathrm{de} g r e e s C}$