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?

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Answer 1 · 2016-03-20T14:02:40

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:

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

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:

$Q_(water)=Q_(object)$

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

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

$c_(object)=(0.75*1*18(cancel(kg)*(kcal)/(cancel(kg)*cancel(C))*cancel(C)))/((0.09*30)(kg*C))$

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

An object with a mass of 90 g90 g is dropped into 750 mL750 mL of water at 0^@C0^@C. If the object cools by 30 ^@C30 ^@C and the water warms by 18 ^@C18 ^@C, what is the specific heat of the material that the object is made of?