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]
Calorimetry
ICSE 2022
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Answer
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 = = 0.168 J g-1 °C-1
Hence, Sp. Heat of metal = 0.168 J g-1 °C-1
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