Question

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?

Answer 1

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_(object)=5(kcal)/(kg*C)`

Explanation:

Known constants:

`c_(water)=1(kcal)/(kg*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_(water)=Q_(object)`

`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_(object)=(0.75*1*18(cancel(kg)*(kcal)/(cancel(kg)*cancel(C))*cancel(C)))/((0.09*30)(kg*C))`

`c_(object)=5(kcal)/(kg*C)`