You make some need tea by dropping 166 g of ice into 500.0 mL of warm tea in an

Question

You make some need tea by dropping 166 g of ice into 500.0 mL of warm tea in an insulated pitcher. If the tea is initially at 18.3 degree C and the ice cubes are initially at 0.00 degree C how many grams of ice will still be present when the contents of the pitcher reach a final temperature? the tea is mostly water, so assume that it has the same density (1.00 g/mL). molar mass, heat capacity (75.3 JK^-1 mol^-1), and heat of fusion (6.00 kJ/mol) as pure water. the heat capacity of ice is 37.7 J K^-1 mol^-1

Explanation / Answer

mass of tea water = 500 g

Q = m Cp dT

    = 500/18 x 75.3 x 18.3

    = 38277.5 J

Q = m L

    = 166 / 18 x 6 x 10^3

    = 55333.3 J

here we need 55333.3 J heat. but we have 38277.5 J

mass = 38277.5 / 333.5 = 114.78 g

so

the mass of ice = 166 - 114.78

mass of ice remaining = 51.2 g

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