An object with a mass of #30 g# is dropped into #50 mL# of water at #0^@C#. If the object cools by #20 ^@C# and the water warms by #80 ^@C#, what is the specific heat of the material that the object is made of?

1 Answer
Andrea B.
Oct 22, 2017

#cp_o = 6,66 (Kcal)/(kg xx °C) = 27,9 (KJ)/(Kg xx K)#

Explanation:

the heat gained from water (50 mL = 50 g) is the same than the heat given from the objec. Hence #M_o xx cp_o xx (Ti-Tf)_o = M_w xx cp_w xx (Tf-Ti)_w#
# 0,030 g xx cp_o xx 20°C = 0,050 kg xx 1( Kcal)/(kg xx °C) xx 80°C#
#0,6 g xx °C xx cp_o = 4kcal#
#cp_o = 6,66 (Kcal)/(kg xx °C) = 27,9 (KJ)/(Kg xx K)#
Pay attention! It doesn't exist any solid object whose specific heat is bigger than water's one
Moreover sperimentally you should considered the effect of the container

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