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Question
40g of ice at 0°C is used to bring down the temperature of a certain mass of water at 60°C to 10°C. Find the mass of water used.
[Specific heat capacity of water = 4200 J kg-1 °C-1]
[Specific latent heat of fusion of ice = 336 × 103 J kg-1]
Solution
Let mass of water used = m gm.
By principal of calorimetry,
Heat given = Heat taken
m1c1ΔQ1 = mL + m2c2ΔQ2
m × 4.2 × (60 - 10) = 40 × 336 + 40 × 4.2 × (10 - 0)
m × 4.2 × 50 = 40 × 336 + 1680
m = 72 gm
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