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प्रश्न
250 g of water at 30 °C is present in a copper vessel of mass 50 g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5 °C.
Specific latent heat of fusion of ice = 336 × 103 J kg-1
Specific heat capacity of copper vessel = 400 J kg-1 °C-1
Specific heat capacity of water = 4200 J kg-1 °C-1.
उत्तर
Given: water = 250 gm = 30 °C
mass of vessel = 50 gm
Final temperature = 5 °C
By the principle of calorimeter.
Heat given = Heat taken,
Let mass of ice = m gm
250 × 4.2 × (30 - 5) + 50 × 0.4 × 25 = m × 336 + m × 4.2 × 5
26250 + 500 = 336 m + 21 m
26750 = 357 m
m = 74.9 gm.
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