# 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?

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).
${M}_{o} C {p}_{o} {\left(T 2 - T 1\right)}_{o} = {M}_{w} C {p}_{w} {\left(T 2 - T 1\right)}_{w}$
since 600 mL of water are 0,6 Kg and $C {p}_{w}$ 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