# An object with a mass of 180 g is dropped into 810 mL of water at 0^@C. If the object cools by 48 ^@C and the water warms by 8 ^@C, what is the specific heat of the material that the object is made of?

Mar 11, 2018

$\text{3.15 J/(g°C)}$

#### Explanation:

Mass of water = Volume × Density $= 810 \cancel{\text{ml" × "1 g"/cancel"ml" = "810 g}}$

$\text{Heat liberated by hot body = -Heat gained by cold body}$

$\text{m"_"o""C"_"o"Δ"T"_"o" = -"m"_"w""C"_"w"Δ"T"_"w}$

"C"_"o" = -("m"_"w""C"_"w"Δ"T"_"w")/("m"_"o"Δ"T"_"o")

$\text{C"_"o" = -("810 g" × "4.2 J/(g°C)" × "8°C")/("180 g" × (-"48°C")) = "3.15 J/(g°C)}$