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
Andrea B.
Apr 27, 2017

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

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