# 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?

##### 1 Answer
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

Answer is:

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

#### 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))

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

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