Question

# 250 g of water at 30℃ is contained in a copper vessel of mass 50g. Calculate the mass of ice required to bring down the temperature of the vessel and its contents to 5℃. Given: specific latent heat of fusion of ice = $$336 ×103J kg^{-1}$$, specific heat capacity of copper = $$400 J kg^{-1} K^{-1}$$, specific heat capacity of water = $$4200 J kg^{-1} K^{-1}$$ .

Solution

## Mass of copper vessel $$m_1 = 50 g$$.Mass of water contained in copper vessel $$m_2 = 250$$g.  Mass of ice required to bring down the temperature of vessel = $$m$$Final temperature =$$5°C$$.Amount of heat gained when 'm' g of ice at 0°C converts into water at 0°C = $$m × 336 J$$Amount of heat gained when temperature of 'm' g of water at 0°C rises to $$5°C = m × 4.2 × 5$$Total amount of heat gained = $$m × 336 + m × 4.2 × 5$$Amount of heat lost when 250 g of water at 30°C cools to $$5°C = 250 × 4.2 × 25 = 26250 J$$ Amount of heat lost when 50 g of vessel at 30°C cools to $$5°C = 50 × 0.4 × 25 = 500 J$$Total amount of heat lost $$= 26250 + 500 = 26750 J$$We know that amount of heat gained = amount of heat lost $$m × 336 + m × 4.2 × 5 = 26750/357$$ $$m =26750/357 = 74.93 g$$Hence, mass of ice required is $$74.93 g$$.Physics

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