Answer:
0.10072 kg
Step-by-step explanation:
To solve this problem, we need to use the heat 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 when cooling the water from 71 F to 32 F. We will assume that the specific heat capacity of water is 4.184 J/(g°C) and the heat of fusion of ice is 333.55 J/g.
Q = (14 oz) x (28.35 g/oz) x (4.184 J/(g°C)) x (71-32) F
Q = 67,576.28 J
To calculate the mass of ice needed, we need to know the final temperature of the water after adding the ice. Assuming all the ice melts and the mixture reaches 32 F, we can use the following equation:
Q = mL + mcΔT
where L is the heat of fusion of ice and m is the mass of ice added.
Substituting the known values, we get:
67,576.28 J = m(333.55 J/g) + (14 oz) x (28.35 g/oz) x (4.184 J/(g°C)) x (71-32) F
Solving for m, we get:
m = 100.72 g or 0.10072 kg
Therefore, we need to add about 0.10072 kg of ice to cool the water from 71 F to 32 F.