# Determine the mass of the remaining ice in the jar. 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...

Determine the mass of the remaining ice in the jar.

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 170 g, 138 g of ice at 0°C is added. At the time at which the temperature of the tea (and melted ice) is 21°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.

Answer in grams.

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### 1 Answer

Depending on the level at which the question is asked there are two possible expected answers: 115 g of ice left and 119 g of ice left. The second is the more accurate answer, but it requires a deeper level of understanding. I've worked out both answers below.

The tea gives heat to the ice, so the amount of heat given by the tea must be equal to the heat taken in by the ice q tea = q water

However, since the tea is giving away heat and the ice is taking it in, the sign on those two values is different (exothermic, giving away heat, is negative) we need to add a (-1) to make them truly equal

(-1)q tea = q water

For a temperature change the formula is q = mcΔT where m is mass, c is the specific heat (4.184 J/g oC for water) and ΔT is the change in temperature (T final – T initial)

For a phase change, the formula is q = mΔHfus where m is the mass and DHfus is the heat of fusion (for ice = 333.55 J/g).

Substituting in gives us:

(-1) m C ΔT (tea) = mΔHfus (ice melting)

(-1)(170g)(4.184 J/g oC)(21-32oC) = (X g)(333.55 J/g)

solving for X gives 23.45 g (23 g with sig figs), meaning that 115 grams of ice are left (138-23).

There is a hitch, however. The ice that melts becomes water at 0oC and must, therefore, have also warmed up (in order for the tea to be at 21oC. So a more complete equation is:

(-1) q tea = q ice melting + q cold water warming

That cold water will be the mass of the melted ice and will be warming from 0oC to 21oC.

Substituting in our values, gives:

(-1)(170 g)(4.184 J/g oC)(21-32oC) = (X g)(333.55 J/g) + (X g)(4.184 J/g oC)(21-0oC)

simplifying, gives us:

(-1)(170 g)(4.184 J/g oC)(21-32oC) = (X g)[333.55 + (4.184*21)]

solving for X gives us 18.57 g ice (19 g with sig figs) meaning that 119 g of ice are left (138-19).