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

Answer:

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)#