Question

A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 30.7 ◦C . In an attempt to cool the liquid, which has a mass of 188 g , 129 g of ice at 0.0 ◦C is added. At the time at which the temperature of the tea is 26.6 ◦C , find the mass of the remaining ice in the jar. The specific heat of water is 4186 J/kg · ◦ C . Assume the specific heat capacity of the tea to be that of pure liquid water. Answer in units of g. (2 significant digits please)

Answer 1

To solve this problem, we need to use the heat transfer equation:

Q = mcΔT

where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

First, let's calculate the heat transferred from the tea to the ice:

Q1 = mcΔT = (0.188 kg)(4186 J/kg·◦C)(30.7 ◦C - 26.6 ◦C) = 342.4 J

This amount of heat is transferred to the ice, causing it to melt and then heat up to the final temperature of the mixture.

Next, let's calculate the heat required to melt the ice:

Q2 = mLf = (0.129 kg)(334 J/g) = 43.14 J

where Lf is the heat of fusion of ice.

Since the heat transferred from the tea (Q1) is greater than the heat required to melt the ice (Q2), all of the ice will melt and then heat up to the final temperature of the mixture.

Finally, let's calculate the mass of the remaining ice:

Q3 = mcΔT = m(4186 J/kg·◦C)(26.6 ◦C - 0.0 ◦C) = 111,483.6 J

This is the amount of heat required to heat up the melted ice to the final temperature of the mixture.

Since the heat transferred from the tea (Q1) is equal to the sum of the heat required to melt the ice (Q2) and the heat required to heat up the melted ice (Q3), we can write:

Q1 = Q2 + Q3

342.4 J = 43.14 J + 111,483.6 J + m(334 J/g)

Solving for m, we get:

m = (342.4 J - 43.14 J - 111,483.6 J) / (334 J/g) = -330.34 g

Since mass cannot be negative, this result means that all of the ice melted and there is no remaining ice in the jar.

Therefore, the mass of the remaining ice is 0 g.