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

Related questions
See all questions in Specific Heat
Impact of this question
1961 views around the world
You can reuse this answer
Creative Commons License