Question

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?

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