# An object with a mass of 160 g is dropped into 750 mL of water at 0^@C. If the object cools by 120 ^@C and the water warms by 5 ^@C, what is the specific heat of the material that the object is made of?

0,203 (Kcal)/(kg°C)= 0,853 (KJ)/(Kg°C)
$M o C p o \left(T o - {T}_{1}\right) = M w C p w \left({T}_{1} - T w\right)$
0,160 kg Cpo (120-5) °C = 0,750 Kg (1(Kcal)/(kg°C)) (5-0)°C
Cpo = (3,75Kcal)/ (18,4 (Kg°C)) = 0,203 (Kcal)/(kg°C)= 0,853 (KJ)/(Kg°C)