An object with a mass of #70 g# is dropped into #600 mL# of water at #0^@C#. If the object cools by #24 ^@C# and the water warms by #8 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Apr 27, 2017

Answer:

#Cp_o = 2,85(Kcal)/(kg°C)#?

Explanation:

the amount of heat changed by the object (Qo) is the same changed by water(Qw).
#M_o Cp_o (T2-T1)_o = M_w Cp_w (T2-T1)_w#
since 600 mL of water are 0,6 Kg and #Cp_w# is 1 Kcal/(kg°C)
#0,07kg Cp_o 24°C = 0,6kg 1(Kcal)/(kg°C) 8°C #
#Cp_o = 2,85(Kcal)/(kg°C)#
the result is impossible because dosen't exist a solid object with a specific heat so high