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?

Question

1542 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-11T16:30:12

$"3.15 J/(g°C)"$

Mass of water = Volume × Density $= 810 cancel"ml" × "1 g"/cancel"ml" = "810 g"$

$"Heat liberated by hot body = -Heat gained by cold body"$

$"m"_"o""C"_"o"Δ"T"_"o" = -"m"_"w""C"_"w"Δ"T"_"w"$

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

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

An object with a mass of 180 g180 g is dropped into 810 mL810 mL of water at 0^@C0^@C. If the object cools by 48 ^@C48 ^@C and the water warms by 8 ^@C8 ^@C, what is the specific heat of the material that the object is made of?