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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