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?

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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)"#