# An object with a mass of 90 g is dropped into 750 mL of water at 0^@C. If the object cools by 30 ^@C and the water warms by 18 ^@C, what is the specific heat of the material that the object is made of?

Mar 20, 2016

Keep in mind that the heat the water receives is equal to the heat that the object loses and that heat is equal to:

Q=m*c*ΔT

${c}_{o b j e c t} = 5 \frac{k c a l}{k g \cdot C}$

#### Explanation:

Known constants:

${c}_{w a t e r} = 1 \frac{k c a l}{k g \cdot C}$

ρ_(water)=1(kg)/(lit)->1kg=1lit which means that litres and kilograms are equal.

The heat that the water received is equal to the heat that the object lost. This heat is equal to:

Q=m*c*ΔT

Therefore:

${Q}_{w a t e r} = {Q}_{o b j e c t}$

m_(water)*c_(water)*ΔT_(water)=m_(object)*color(green)(c_(object))*ΔT_(object)

c_(object)=(m_(water)*c_(water)*ΔT_(water))/(m_(object)*ΔT_(object))

${c}_{o b j e c t} = \frac{0.75 \cdot 1 \cdot 18 \left(\cancel{k g} \cdot \frac{k c a l}{\cancel{k g} \cdot \cancel{C}} \cdot \cancel{C}\right)}{\left(0.09 \cdot 30\right) \left(k g \cdot C\right)}$

${c}_{o b j e c t} = 5 \frac{k c a l}{k g \cdot C}$