A 140.0-g sample of water at 25.0°c is mixed with 111.7 g of a certain metal at 100.0°c. after thermal equilibrium is established, the (final) temperature of the mixture is 29.6°c. what is the specific heat capacity of the metal, assuming it is constant over the temperature range concerned?

Question

Chemistry

Jaylan Travis

15 February, 00:17

A 140.0-g sample of water at 25.0°c is mixed with 111.7 g of a certain metal at 100.0°c. after thermal equilibrium is established, the (final) temperature of the mixture is 29.6°c. what is the specific heat capacity of the metal, assuming it is constant over the temperature range concerned?

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Warren Hunter · Answer 1

Mass of water = 140.0 g

Initial temperature of water = 25.0°C

Mass of a certain metal = 111.7 g

Initial temperature of metal = 100.0°C

Final temperature of water and metal = 29.6°C

Since the metal is at a higher initial temperature it will lose heat and the water having a lower initial temperature will gain heat.

Thus, heat lost by metal = Heat gained by water

formula: (mass metal) (initialT - finalT) (Cp metal) = (mass water) (finalT - initalT) (Cp water)

After plugging in the given data we get,

(111.7g) (100 °C - 29.6°C) (Cp metal) = (140.0g) (29.6°C-25.0°C) (4.184 J/g°C)

(111.7g) (70.4°C) (Cp metal) = (140.0g) (4.6°C) (4.184 J/g°C)

(7863.7 g°C) (Cp metal) = 2694.5 J

(Cp metal) = 2694.5 J / 7863.7 g°C

Thus, Cp of metal = 0.3427 J/g°C