A person makes ice tea by adding ice to 1.8 kg of hot tea,initially at 80°C.How many kilograms of ice,initially at 0.00°C,are required to bring the mixture to 10°C? The heat of fusion of ice is 334 kJ/kg,and we can assume that tea has essentially the same thermal properties as water,so its specific heat is 4190 J/(kg • K).
A) 1.0 kg
B) 1.2 kg
C) 1.4 kg
D) 1.5 kg
E) 1.7 kg

Question

Quizplus · Accepted Answer

First, we need to calculate the amount of thermal energy released by the tea as it cools from 80°C to 10°C. Using the specific heat of water/tea, we have:

Q1 = m1c∆T
Q1 = 1.8 kg x 4190 J/(kg • K) x (80°C - 10°C)
Q1 = 1.8 kg x 4190 J/(kg • K) x 70°C
Q1 = 526920 J

This energy will be used to melt ice, so we need to calculate how much ice is required to absorb this amount of energy. The heat of fusion of ice is given as 334 kJ/kg, which can be converted to J/kg:

334 kJ/kg x 1000 J/kJ = 334000 J/kg

Using Q = ml, where Q is the thermal energy absorbed/released, m is the mass of the substance, and l is the latent heat of fusion:

Q2 = m2l
m2 = Q2/l
m2 = 526920 J/334000 J/kg
m2 = 1.575 kg

Therefore, we need 1.575 kg of ice to absorb the thermal energy released by the tea as it cools. However, this much ice would make the mixture colder than 10°C, so we need to add more ice to bring the temperature up to 10°C. The amount of thermal energy required to cool the mixture from 0°C to 10°C can be calculated as:

Q3 = m3c∆T
Q3 = m3 x 4190 J/(kg • K) x 10°C
Q3 = 41900m3 J

Since this energy needs to be removed from the tea/ice mixture, we can set up an equation:

Q1 + Q2 + Q3 = 0

Substituting the values we've calculated:

526920 J + 1.575 kg x 334000 J/kg + 41900m3 J = 0
1.575 kg x 334000 J/kg + 41900m3 J = -526920 J
1.575 kg x 334000 J/kg = -526920 J - 41900m3 J
523950 J = -41900m3 J
m3 = -523950 J/-41900 J/kg
m3 = 12.5 kg

Since we need to add ice to the mixture, we take the absolute value of m3:

m3 = 12.5 kg

Finally, we add the amount of ice required to absorb the thermal energy released by the tea and the amount of ice required to bring the mixture to 10°C:

m_total = m2 + m3
m_total = 1.575 kg + 12.5 kg
m_total = 14.075 kg ≈ 1.4 kg

Therefore, the answer is (C) 1.4 kg of ice.

A Person Makes Ice Tea by Adding Ice to 1