Physics

A metal piece of mass 420 g present at 80°C is dropped in 80 g of water present at 20°C in a calorimeter of mass 84 g. If the final temperature of the mixture is 30°C, then calculate the specific heat capacity of the metal piece.
[Specific heat capacity of water = 4.2 Jg^-1 °C^-1, Specific heat capacity of the calorimeter = 200 Jkg^-1 °C^-1]

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Given,

Mass of metal = 420 g at temperature = 80°C

Mass of water = 80 g at temperature = 20°C

Sp. Heat of water = 4.2 J g^-1 °C^-1

Mass of calorimeter = 84 g

Sp. heat of calorimeter = 200 J kg^-1 °C^-1 = 0.2 J g^-1 °C^-1

Final temperature of mixture = 30°C

Sp. Heat of metal [S] = ?

Heat energy lost by metal
= mass of metal x specific heat capacity x fall in temperature
= 420 x S x (80-30)
= 420 x S x 50
= 21000 S J

Heat energy gained by water
= mass of water x specific heat capacity x rise in temperature
= 80 x 4.2 x (30-20)
= 80 x 4.2 x 10
= 3360 J

Heat energy gained by calorimeter
= mass of calorimeter x specific heat capacity x rise in temperature
= 84 x 0.2 x (30-20)
= 84 x 0.2 x 10
= 168 J

As, heat energy lost = heat energy gained

∴ 21000S = 3360 + 168

∴ S = $\dfrac{3528}{21000}$ = 0.168 J g^-1 °C^-1

Hence, Sp. Heat of metal = 0.168 J g^-1 °C^-1

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