The minimum mass of ice required to cool 250 mL of coffee from 90°C to 0°C is 250 g.
To calculate the minimum mass of ice required to cool 250 mL of coffee from 90°C to 0°C, we can use the equation:
Q = mcΔT
Where Q is the amount of heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
Since the coffee is cooling down, the heat transferred will be negative:
Q = -mcΔT
We can rearrange the equation to solve for m:
m = Q / (cΔT)
In this case, the coffee is going from 90°C to 0°C, so ΔT = -90°C.
The specific heat capacity of coffee is similar to water, which is about 4.184 J/(g•°C).
By using the given data, we can calculate the mass:
m = Q / (4.184 J/(g•°C) * -90°C)
Now, we can convert the volume of coffee to mass by considering its density. Since the density of coffee is about 1.00 g/mL, the mass of 250 mL of coffee is 250 g.
Therefore, the minimum mass of ice required to cool 250 mL of coffee from 90°C to 0°C is 250 g. Answer choice (c) is correct.