# What mass of ice remains once the liquid cools to 12°C?A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 32°C. In an attempt to cool the liquid,which has a mass of...

What mass of ice remains once the liquid cools to 12°C?

A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 32°C.

In an attempt to cool the liquid,which has a mass of 155 g, 131 g of ice at 0°C is added.At the time at which the temperature of the tea (and melted ice) is 12°C, determine the mass of the remaining ice in the jar. Assume the specific heat capacity of the tea to be that of pure liquid water.

mlsiasebs | Certified Educator

Heat is transferred from the tea to the ice.  As a result, the tea cools and the energy transferred to the ice is melted.  By finding the energy change which results in the cooling of the tea, we can find out how much ice will melt with that amount of energy.  The difference in the amount of ice we had originally and the amount that melts will be the mass of ice that remains.

To determine the heat change in the tea, we need to use the formula

q = m * s * deltaT (where m is the mass of the liquid and s is the specific heat)

For water, s is the specific heat of the tea for which we can use the specific heat of water, 4.184 J/g°C

q = (155 g)(4.184 J/g°C)(32°C-12°C)

q = 12970 J = 12.97 kJ

This is the amount of energy transferred from the liquid which melts the ice.  By using the enthalpy of fusion for the ice, we can determine the amount of ice that will melt.

The enthalpy of fusion for ice is 6.01 kJ/mol.  We can use the formula

q = n * deltaH where n is the moles of ice and deltaH is the enthalpy of fusion of ice.

12.97 kJ = n (6.01 kJ/mol)

Solving for n, we find

n = 2.16 moles

which we can convert to grams using the molar mass of water.

2.16 moles (18.01 g/mol) = 38.9 g of ice needed to cool the tea.

131 g of ice originally present - 38.9 g of ice used = 92 g of ice remaining

najm1947 | Student

Initial Temperature of Liquid (Tea) = 32 deg C

Mass of Tea = 155 g

Final Temperature of Liquid (Tea + Molted Ice) = 12 deg C

Specific Heat of Water/Tea = 4.186 J/g deg C

Latent Heat (Heat of Fusion) of Ice = 333.55 J/g

Original Mass of Ice = 131 g

Heat Lost/Gained by Liquid = s*m*T where s is specific heat, m is mass and T is the change in temperature

Heat lost by Tea in cooling = 4.186*155*(32-12) = 14923 J

Let mi be the mass of Ice in grams that melted in to water

Heat Gained by Ice in melting = Latent Heat * mass = 333.55*mi J

Heat gained by Melted Ice from Temperature Rise from 0 to 12 deg C = 4.186*mi*(12-0) = 50.23*mi J

Total Heat Gain = 333.55*mi + 50.23*mi = 383.78*mi J

Heat Gain = Heat Loss for the system

383.78*mi = 14923

mi = 38.9 g

The mass of Ice melted = 38.9 g

The mass of Remaining Ice in Jar = 131 - 38.9 = 92.1 g