A 50.0 g piece of metal at 100.0 degree C was plunged into 100.0 g of water at 24.0 degree C. The temperature of the resulting mixture became 28.0 degree C. (a) How many joules did the water absorbs (b) How many joules did the metal lose?

1 Answer
Ernest Z.
Mar 23, 2015

(a) The water absorbed 1670 J.
(b) The metal lost 1670 J.
(c) The heat capacity of the metal is 420 J/°C.
(d) The specific heat capacity of the metal is 8.4 J·°C⁻¹g⁻¹.

(a) Heat absorbed by the water

#ΔT = (28.0 – 24.0)°C = 4.0 °C#

#q_"H₂O" = mcΔT = 100.0 cancel("g") × "4.184 J·"cancel("°C⁻¹g⁻¹") × 4.0 cancel("°C") = "1670 J"#

(b) Heat lost by the metal

Heat lost by metal = heat gained by water = 1670 J

(c) Heat capacity of the metal

#q_"metal" = CΔT#

#C = q_"metal"/(ΔT) = "1670 J"/"4.0 °C" = "420 J/°C"#

(d) Specific heat capacity of the metal

Specific heat capacity means the heat capacity per gram.

#c = C/m = ("420 J·°C"^-1)/"50.0 g" = "8.4 J·°C"^-1"g"^-1"#

Note: The answers can have only 2 significant figures, because that is all you had for the temperature change. If you need more precision, you will have to recalculate.

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